Capacitors with Dielectric
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359207 A parallel plate capacitor has plates of area \(A\) separted by distance \(‘d’\) between them. It is filled with a dielectric which has a dielectric constant that varies as \(k\left( x \right) = K\left( {1 + \alpha x} \right)\) where \(‘x’\) is the distance measured from one of the plates. If \(\left( {\alpha d} \right) < < 1\), the total capacitance of the system is best given by the expression :
supporting img

1 \(\frac{{AK{ \in _0}}}{d}\left( {1 + \alpha d} \right)\)
2 \(\frac{{A{ \in _0}K}}{d}\left( {1 + \frac{{{\alpha ^2}{d^2}}}{2}} \right)\)
3 \(\frac{{AK{ \in _0}}}{d}\left( {1 + \frac{{\alpha d}}{2}} \right)\)
4 \(\frac{{A{ \in _0}K}}{d}\left( {1 + {{\left( {\frac{{\alpha d}}{2}} \right)}^2}} \right)\)
JEE - 2020
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359208 In the figure a capacitor is filled with dielectrics. The resultant capacitance is
supporting img

1 \(\frac{{{\varepsilon _o}A}}{d}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}} \right]\)
2 \(\frac{{2{\varepsilon _o}A}}{d}\left[ {{k_1} + {k_2} + {k_3}} \right]\)
3 \(\frac{{2{\varepsilon _o}A}}{d}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}} \right]\)
4 None of these
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359209 Assertion :
A parallel plate capacitor is charged to a potential difference of \(100\;V\), and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Then values of capacitance increases.
Reason :
When we insert a dielectric between the plates of a capacitor. Potential energy stored in the capacitor decreases

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359210 A parallel plate air capacitor has a capacitance \(C\). When it is half filled with dielectric of dielectric constant 5, the percentage increase in the capacitance will be

1 \(400\% \)
2 \(66.6\% \)
3 \(33.3\% \)
4 \(200\% \)
KCET - 2006
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359211 A capacitor is formed by two square metal-plates of edge \(a\) separated by a distance \(d\). Dielectrics of dielectric constants \({K_1}\) and \({K_2}\) are filled in the gap as shown in figure. Find the capacitance.
supporting img

1 \(\frac{{{\varepsilon _0}{K_1}{K_2}}}{{{K_1} - {K_2}}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
2 \(\frac{{{\varepsilon _0}{K_1}{K_2}a}}{{\left( {{K_1} - {K_2}} \right)}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
3 \(\frac{{{\varepsilon _0}{K_1}{K_2}{a^3}}}{{{K_1} - {K_2}}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
4 \(\frac{{{\varepsilon _0}{K_1}{K_2}{a^2}}}{{d\left( {{K_1} - {K_2}} \right)}}\,ln\left( {\frac{{{K_1}}}{{{k_2}}}} \right)\)
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359207 A parallel plate capacitor has plates of area \(A\) separted by distance \(‘d’\) between them. It is filled with a dielectric which has a dielectric constant that varies as \(k\left( x \right) = K\left( {1 + \alpha x} \right)\) where \(‘x’\) is the distance measured from one of the plates. If \(\left( {\alpha d} \right) < < 1\), the total capacitance of the system is best given by the expression :
supporting img

1 \(\frac{{AK{ \in _0}}}{d}\left( {1 + \alpha d} \right)\)
2 \(\frac{{A{ \in _0}K}}{d}\left( {1 + \frac{{{\alpha ^2}{d^2}}}{2}} \right)\)
3 \(\frac{{AK{ \in _0}}}{d}\left( {1 + \frac{{\alpha d}}{2}} \right)\)
4 \(\frac{{A{ \in _0}K}}{d}\left( {1 + {{\left( {\frac{{\alpha d}}{2}} \right)}^2}} \right)\)
JEE - 2020
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359208 In the figure a capacitor is filled with dielectrics. The resultant capacitance is
supporting img

1 \(\frac{{{\varepsilon _o}A}}{d}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}} \right]\)
2 \(\frac{{2{\varepsilon _o}A}}{d}\left[ {{k_1} + {k_2} + {k_3}} \right]\)
3 \(\frac{{2{\varepsilon _o}A}}{d}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}} \right]\)
4 None of these
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359209 Assertion :
A parallel plate capacitor is charged to a potential difference of \(100\;V\), and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Then values of capacitance increases.
Reason :
When we insert a dielectric between the plates of a capacitor. Potential energy stored in the capacitor decreases

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359210 A parallel plate air capacitor has a capacitance \(C\). When it is half filled with dielectric of dielectric constant 5, the percentage increase in the capacitance will be

1 \(400\% \)
2 \(66.6\% \)
3 \(33.3\% \)
4 \(200\% \)
KCET - 2006
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359211 A capacitor is formed by two square metal-plates of edge \(a\) separated by a distance \(d\). Dielectrics of dielectric constants \({K_1}\) and \({K_2}\) are filled in the gap as shown in figure. Find the capacitance.
supporting img

1 \(\frac{{{\varepsilon _0}{K_1}{K_2}}}{{{K_1} - {K_2}}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
2 \(\frac{{{\varepsilon _0}{K_1}{K_2}a}}{{\left( {{K_1} - {K_2}} \right)}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
3 \(\frac{{{\varepsilon _0}{K_1}{K_2}{a^3}}}{{{K_1} - {K_2}}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
4 \(\frac{{{\varepsilon _0}{K_1}{K_2}{a^2}}}{{d\left( {{K_1} - {K_2}} \right)}}\,ln\left( {\frac{{{K_1}}}{{{k_2}}}} \right)\)
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359207 A parallel plate capacitor has plates of area \(A\) separted by distance \(‘d’\) between them. It is filled with a dielectric which has a dielectric constant that varies as \(k\left( x \right) = K\left( {1 + \alpha x} \right)\) where \(‘x’\) is the distance measured from one of the plates. If \(\left( {\alpha d} \right) < < 1\), the total capacitance of the system is best given by the expression :
supporting img

1 \(\frac{{AK{ \in _0}}}{d}\left( {1 + \alpha d} \right)\)
2 \(\frac{{A{ \in _0}K}}{d}\left( {1 + \frac{{{\alpha ^2}{d^2}}}{2}} \right)\)
3 \(\frac{{AK{ \in _0}}}{d}\left( {1 + \frac{{\alpha d}}{2}} \right)\)
4 \(\frac{{A{ \in _0}K}}{d}\left( {1 + {{\left( {\frac{{\alpha d}}{2}} \right)}^2}} \right)\)
JEE - 2020
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359208 In the figure a capacitor is filled with dielectrics. The resultant capacitance is
supporting img

1 \(\frac{{{\varepsilon _o}A}}{d}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}} \right]\)
2 \(\frac{{2{\varepsilon _o}A}}{d}\left[ {{k_1} + {k_2} + {k_3}} \right]\)
3 \(\frac{{2{\varepsilon _o}A}}{d}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}} \right]\)
4 None of these
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359209 Assertion :
A parallel plate capacitor is charged to a potential difference of \(100\;V\), and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Then values of capacitance increases.
Reason :
When we insert a dielectric between the plates of a capacitor. Potential energy stored in the capacitor decreases

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359210 A parallel plate air capacitor has a capacitance \(C\). When it is half filled with dielectric of dielectric constant 5, the percentage increase in the capacitance will be

1 \(400\% \)
2 \(66.6\% \)
3 \(33.3\% \)
4 \(200\% \)
KCET - 2006
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359211 A capacitor is formed by two square metal-plates of edge \(a\) separated by a distance \(d\). Dielectrics of dielectric constants \({K_1}\) and \({K_2}\) are filled in the gap as shown in figure. Find the capacitance.
supporting img

1 \(\frac{{{\varepsilon _0}{K_1}{K_2}}}{{{K_1} - {K_2}}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
2 \(\frac{{{\varepsilon _0}{K_1}{K_2}a}}{{\left( {{K_1} - {K_2}} \right)}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
3 \(\frac{{{\varepsilon _0}{K_1}{K_2}{a^3}}}{{{K_1} - {K_2}}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
4 \(\frac{{{\varepsilon _0}{K_1}{K_2}{a^2}}}{{d\left( {{K_1} - {K_2}} \right)}}\,ln\left( {\frac{{{K_1}}}{{{k_2}}}} \right)\)
NEET DIGITAL LIBRARY
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359207 A parallel plate capacitor has plates of area \(A\) separted by distance \(‘d’\) between them. It is filled with a dielectric which has a dielectric constant that varies as \(k\left( x \right) = K\left( {1 + \alpha x} \right)\) where \(‘x’\) is the distance measured from one of the plates. If \(\left( {\alpha d} \right) < < 1\), the total capacitance of the system is best given by the expression :
supporting img

1 \(\frac{{AK{ \in _0}}}{d}\left( {1 + \alpha d} \right)\)
2 \(\frac{{A{ \in _0}K}}{d}\left( {1 + \frac{{{\alpha ^2}{d^2}}}{2}} \right)\)
3 \(\frac{{AK{ \in _0}}}{d}\left( {1 + \frac{{\alpha d}}{2}} \right)\)
4 \(\frac{{A{ \in _0}K}}{d}\left( {1 + {{\left( {\frac{{\alpha d}}{2}} \right)}^2}} \right)\)
JEE - 2020
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359208 In the figure a capacitor is filled with dielectrics. The resultant capacitance is
supporting img

1 \(\frac{{{\varepsilon _o}A}}{d}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}} \right]\)
2 \(\frac{{2{\varepsilon _o}A}}{d}\left[ {{k_1} + {k_2} + {k_3}} \right]\)
3 \(\frac{{2{\varepsilon _o}A}}{d}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}} \right]\)
4 None of these
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359209 Assertion :
A parallel plate capacitor is charged to a potential difference of \(100\;V\), and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Then values of capacitance increases.
Reason :
When we insert a dielectric between the plates of a capacitor. Potential energy stored in the capacitor decreases

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359210 A parallel plate air capacitor has a capacitance \(C\). When it is half filled with dielectric of dielectric constant 5, the percentage increase in the capacitance will be

1 \(400\% \)
2 \(66.6\% \)
3 \(33.3\% \)
4 \(200\% \)
KCET - 2006
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359211 A capacitor is formed by two square metal-plates of edge \(a\) separated by a distance \(d\). Dielectrics of dielectric constants \({K_1}\) and \({K_2}\) are filled in the gap as shown in figure. Find the capacitance.
supporting img

1 \(\frac{{{\varepsilon _0}{K_1}{K_2}}}{{{K_1} - {K_2}}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
2 \(\frac{{{\varepsilon _0}{K_1}{K_2}a}}{{\left( {{K_1} - {K_2}} \right)}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
3 \(\frac{{{\varepsilon _0}{K_1}{K_2}{a^3}}}{{{K_1} - {K_2}}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
4 \(\frac{{{\varepsilon _0}{K_1}{K_2}{a^2}}}{{d\left( {{K_1} - {K_2}} \right)}}\,ln\left( {\frac{{{K_1}}}{{{k_2}}}} \right)\)
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359207 A parallel plate capacitor has plates of area \(A\) separted by distance \(‘d’\) between them. It is filled with a dielectric which has a dielectric constant that varies as \(k\left( x \right) = K\left( {1 + \alpha x} \right)\) where \(‘x’\) is the distance measured from one of the plates. If \(\left( {\alpha d} \right) < < 1\), the total capacitance of the system is best given by the expression :
supporting img

1 \(\frac{{AK{ \in _0}}}{d}\left( {1 + \alpha d} \right)\)
2 \(\frac{{A{ \in _0}K}}{d}\left( {1 + \frac{{{\alpha ^2}{d^2}}}{2}} \right)\)
3 \(\frac{{AK{ \in _0}}}{d}\left( {1 + \frac{{\alpha d}}{2}} \right)\)
4 \(\frac{{A{ \in _0}K}}{d}\left( {1 + {{\left( {\frac{{\alpha d}}{2}} \right)}^2}} \right)\)
JEE - 2020
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359208 In the figure a capacitor is filled with dielectrics. The resultant capacitance is
supporting img

1 \(\frac{{{\varepsilon _o}A}}{d}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}} \right]\)
2 \(\frac{{2{\varepsilon _o}A}}{d}\left[ {{k_1} + {k_2} + {k_3}} \right]\)
3 \(\frac{{2{\varepsilon _o}A}}{d}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}} \right]\)
4 None of these
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359209 Assertion :
A parallel plate capacitor is charged to a potential difference of \(100\;V\), and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Then values of capacitance increases.
Reason :
When we insert a dielectric between the plates of a capacitor. Potential energy stored in the capacitor decreases

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359210 A parallel plate air capacitor has a capacitance \(C\). When it is half filled with dielectric of dielectric constant 5, the percentage increase in the capacitance will be

1 \(400\% \)
2 \(66.6\% \)
3 \(33.3\% \)
4 \(200\% \)
KCET - 2006
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359211 A capacitor is formed by two square metal-plates of edge \(a\) separated by a distance \(d\). Dielectrics of dielectric constants \({K_1}\) and \({K_2}\) are filled in the gap as shown in figure. Find the capacitance.
supporting img

1 \(\frac{{{\varepsilon _0}{K_1}{K_2}}}{{{K_1} - {K_2}}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
2 \(\frac{{{\varepsilon _0}{K_1}{K_2}a}}{{\left( {{K_1} - {K_2}} \right)}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
3 \(\frac{{{\varepsilon _0}{K_1}{K_2}{a^3}}}{{{K_1} - {K_2}}}\,ln\left( {\frac{{{K_2}}}{{{k_1}}}} \right)\)
4 \(\frac{{{\varepsilon _0}{K_1}{K_2}{a^2}}}{{d\left( {{K_1} - {K_2}} \right)}}\,ln\left( {\frac{{{K_1}}}{{{k_2}}}} \right)\)