358087
Two charges are at a distance d apart. If a copper plate of thickness \(\frac{d}{2}\) is kept between them, the effective force will be
1 \(F/2\)
2 \({\rm{zero}}\)
3 \(2F\)
4 \(\sqrt 2 F\)
Explanation:
The dielectric constant for metal is infinity, the force between the two charges would be reduced to zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358088
Force between two point charges \(q_{1}\) and \(q_{2}\) placed in vacuum at '\(r\)' \(cm\) apart is \(F\). Force between them when placed in a medium having dielectric constant \(k=5\) at '\(r / 5\)' \(cm\) apart will be
1 \(25 F\)
2 \(F / 5\)
3 \(5 F\)
4 \(F / 25\)
Explanation:
Given: \(F = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}\quad \quad (1)\) When point charges \(q_{1}\) and \(q_{2}\) placed in medium with dielectric constant \(k=5\) and \(r' = \frac{r}{5},{\rm{ then }}\) \(F' = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{{r'}^2}k}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0} \times 5{{\left( {\frac{r}{5}} \right)}^2}}} = 5\,F\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358089
Two identical charges repel each other with a force equal to \(10\,g\,wt\) when they are \(0.9\,m\) apart in air.\((g = 10m{s^2})\).The value of each charge is
358090
Force of attraction between two point electric charges placed at a distance \(d\) in a medium is \(F\). What distance apart should these be kept in the same medium, so that force between them becomes \(F/3\)?
1 \(\sqrt 3 \,d\)
2 \(2\sqrt 3 \,d\)
3 \(3{\kern 1pt} d\)
4 \(9{\kern 1pt} d\)
Explanation:
Let \({q_1}\) and \({q_2}\) be the two point charges. The force between the charges, at a separation of \(d\). \(F = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{d^2}}}\) Suppose that force between the two charges become \(F/3\), when the charges are kept at a distance apart. Then \(\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{x^2}}} = \frac{F}{3}\) or \(\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{x^2}}} = \frac{1}{3}\left( {\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{d^2}}}} \right)\) or \(x = \sqrt 3 \,d\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII01:ELECTRIC CHARGES AND FIELDS
358087
Two charges are at a distance d apart. If a copper plate of thickness \(\frac{d}{2}\) is kept between them, the effective force will be
1 \(F/2\)
2 \({\rm{zero}}\)
3 \(2F\)
4 \(\sqrt 2 F\)
Explanation:
The dielectric constant for metal is infinity, the force between the two charges would be reduced to zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358088
Force between two point charges \(q_{1}\) and \(q_{2}\) placed in vacuum at '\(r\)' \(cm\) apart is \(F\). Force between them when placed in a medium having dielectric constant \(k=5\) at '\(r / 5\)' \(cm\) apart will be
1 \(25 F\)
2 \(F / 5\)
3 \(5 F\)
4 \(F / 25\)
Explanation:
Given: \(F = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}\quad \quad (1)\) When point charges \(q_{1}\) and \(q_{2}\) placed in medium with dielectric constant \(k=5\) and \(r' = \frac{r}{5},{\rm{ then }}\) \(F' = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{{r'}^2}k}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0} \times 5{{\left( {\frac{r}{5}} \right)}^2}}} = 5\,F\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358089
Two identical charges repel each other with a force equal to \(10\,g\,wt\) when they are \(0.9\,m\) apart in air.\((g = 10m{s^2})\).The value of each charge is
358090
Force of attraction between two point electric charges placed at a distance \(d\) in a medium is \(F\). What distance apart should these be kept in the same medium, so that force between them becomes \(F/3\)?
1 \(\sqrt 3 \,d\)
2 \(2\sqrt 3 \,d\)
3 \(3{\kern 1pt} d\)
4 \(9{\kern 1pt} d\)
Explanation:
Let \({q_1}\) and \({q_2}\) be the two point charges. The force between the charges, at a separation of \(d\). \(F = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{d^2}}}\) Suppose that force between the two charges become \(F/3\), when the charges are kept at a distance apart. Then \(\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{x^2}}} = \frac{F}{3}\) or \(\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{x^2}}} = \frac{1}{3}\left( {\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{d^2}}}} \right)\) or \(x = \sqrt 3 \,d\)
358087
Two charges are at a distance d apart. If a copper plate of thickness \(\frac{d}{2}\) is kept between them, the effective force will be
1 \(F/2\)
2 \({\rm{zero}}\)
3 \(2F\)
4 \(\sqrt 2 F\)
Explanation:
The dielectric constant for metal is infinity, the force between the two charges would be reduced to zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358088
Force between two point charges \(q_{1}\) and \(q_{2}\) placed in vacuum at '\(r\)' \(cm\) apart is \(F\). Force between them when placed in a medium having dielectric constant \(k=5\) at '\(r / 5\)' \(cm\) apart will be
1 \(25 F\)
2 \(F / 5\)
3 \(5 F\)
4 \(F / 25\)
Explanation:
Given: \(F = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}\quad \quad (1)\) When point charges \(q_{1}\) and \(q_{2}\) placed in medium with dielectric constant \(k=5\) and \(r' = \frac{r}{5},{\rm{ then }}\) \(F' = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{{r'}^2}k}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0} \times 5{{\left( {\frac{r}{5}} \right)}^2}}} = 5\,F\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358089
Two identical charges repel each other with a force equal to \(10\,g\,wt\) when they are \(0.9\,m\) apart in air.\((g = 10m{s^2})\).The value of each charge is
358090
Force of attraction between two point electric charges placed at a distance \(d\) in a medium is \(F\). What distance apart should these be kept in the same medium, so that force between them becomes \(F/3\)?
1 \(\sqrt 3 \,d\)
2 \(2\sqrt 3 \,d\)
3 \(3{\kern 1pt} d\)
4 \(9{\kern 1pt} d\)
Explanation:
Let \({q_1}\) and \({q_2}\) be the two point charges. The force between the charges, at a separation of \(d\). \(F = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{d^2}}}\) Suppose that force between the two charges become \(F/3\), when the charges are kept at a distance apart. Then \(\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{x^2}}} = \frac{F}{3}\) or \(\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{x^2}}} = \frac{1}{3}\left( {\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{d^2}}}} \right)\) or \(x = \sqrt 3 \,d\)
358087
Two charges are at a distance d apart. If a copper plate of thickness \(\frac{d}{2}\) is kept between them, the effective force will be
1 \(F/2\)
2 \({\rm{zero}}\)
3 \(2F\)
4 \(\sqrt 2 F\)
Explanation:
The dielectric constant for metal is infinity, the force between the two charges would be reduced to zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358088
Force between two point charges \(q_{1}\) and \(q_{2}\) placed in vacuum at '\(r\)' \(cm\) apart is \(F\). Force between them when placed in a medium having dielectric constant \(k=5\) at '\(r / 5\)' \(cm\) apart will be
1 \(25 F\)
2 \(F / 5\)
3 \(5 F\)
4 \(F / 25\)
Explanation:
Given: \(F = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}\quad \quad (1)\) When point charges \(q_{1}\) and \(q_{2}\) placed in medium with dielectric constant \(k=5\) and \(r' = \frac{r}{5},{\rm{ then }}\) \(F' = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{{r'}^2}k}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0} \times 5{{\left( {\frac{r}{5}} \right)}^2}}} = 5\,F\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358089
Two identical charges repel each other with a force equal to \(10\,g\,wt\) when they are \(0.9\,m\) apart in air.\((g = 10m{s^2})\).The value of each charge is
358090
Force of attraction between two point electric charges placed at a distance \(d\) in a medium is \(F\). What distance apart should these be kept in the same medium, so that force between them becomes \(F/3\)?
1 \(\sqrt 3 \,d\)
2 \(2\sqrt 3 \,d\)
3 \(3{\kern 1pt} d\)
4 \(9{\kern 1pt} d\)
Explanation:
Let \({q_1}\) and \({q_2}\) be the two point charges. The force between the charges, at a separation of \(d\). \(F = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{d^2}}}\) Suppose that force between the two charges become \(F/3\), when the charges are kept at a distance apart. Then \(\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{x^2}}} = \frac{F}{3}\) or \(\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{x^2}}} = \frac{1}{3}\left( {\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{{q_1}\,{q_2}}}{{{d^2}}}} \right)\) or \(x = \sqrt 3 \,d\)