272290
Seven capacitors each of capacitance \(2 \mu F\) are to be connected in a configuration to obtain an effective capacitance of \(\left(\frac{10}{11}\right) \mu F\). Which of the combination \(\{5\}\) shown in figure will achieve the desired result?
1
2
3
4
Explanation:
(a) The equivalent capacitance
NCERT Page-78
Electrostatic Potentials and Capacitance
272291
A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be:
1 5 N
2 10 N
3 20 N
4 Zero
Explanation:
(a) Force on charge particle will be given by
\(\begin{aligned}
& F=q E \\
& F \propto E \\
& F_2=F_1\left(\frac{E_2}{E_1}\right)=F_1\left(\frac{\frac{\sigma}{2 \mathrm{E}_0}}{\frac{\sigma}{E_0}}\right)=\frac{F_1}{2}=\frac{10}{2}=5 \mathrm{~N}
\end{aligned}\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272292
A number of capacitors each of equal capacitance \(C\), are arranged as shown in Fig. The equivalent capacitance between \(A\) and \(B\) is
272293
A capacitor of capacity \(C_1\) is chargedupto \(V\) volt and then connected to an uncharged capacitor of capacity \(C_2\). Then final potential difference across each will be
1 \(\frac{C_2 V}{C_1+C_2}\)
2 \(\left(1+\frac{C_2}{C_1}\right) V\)
3 \(\frac{c_1 V}{c_1+c_2}\)
4 \(\left(1-\frac{c_2}{c_1}\right) V\)
Explanation:
(c) Common potential \(V=\frac{c_1 V+c_2 \times 0}{c_1+c_2}=\frac{c_1}{c_1+c_2} V\)
272290
Seven capacitors each of capacitance \(2 \mu F\) are to be connected in a configuration to obtain an effective capacitance of \(\left(\frac{10}{11}\right) \mu F\). Which of the combination \(\{5\}\) shown in figure will achieve the desired result?
1
2
3
4
Explanation:
(a) The equivalent capacitance
NCERT Page-78
Electrostatic Potentials and Capacitance
272291
A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be:
1 5 N
2 10 N
3 20 N
4 Zero
Explanation:
(a) Force on charge particle will be given by
\(\begin{aligned}
& F=q E \\
& F \propto E \\
& F_2=F_1\left(\frac{E_2}{E_1}\right)=F_1\left(\frac{\frac{\sigma}{2 \mathrm{E}_0}}{\frac{\sigma}{E_0}}\right)=\frac{F_1}{2}=\frac{10}{2}=5 \mathrm{~N}
\end{aligned}\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272292
A number of capacitors each of equal capacitance \(C\), are arranged as shown in Fig. The equivalent capacitance between \(A\) and \(B\) is
272293
A capacitor of capacity \(C_1\) is chargedupto \(V\) volt and then connected to an uncharged capacitor of capacity \(C_2\). Then final potential difference across each will be
1 \(\frac{C_2 V}{C_1+C_2}\)
2 \(\left(1+\frac{C_2}{C_1}\right) V\)
3 \(\frac{c_1 V}{c_1+c_2}\)
4 \(\left(1-\frac{c_2}{c_1}\right) V\)
Explanation:
(c) Common potential \(V=\frac{c_1 V+c_2 \times 0}{c_1+c_2}=\frac{c_1}{c_1+c_2} V\)
272290
Seven capacitors each of capacitance \(2 \mu F\) are to be connected in a configuration to obtain an effective capacitance of \(\left(\frac{10}{11}\right) \mu F\). Which of the combination \(\{5\}\) shown in figure will achieve the desired result?
1
2
3
4
Explanation:
(a) The equivalent capacitance
NCERT Page-78
Electrostatic Potentials and Capacitance
272291
A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be:
1 5 N
2 10 N
3 20 N
4 Zero
Explanation:
(a) Force on charge particle will be given by
\(\begin{aligned}
& F=q E \\
& F \propto E \\
& F_2=F_1\left(\frac{E_2}{E_1}\right)=F_1\left(\frac{\frac{\sigma}{2 \mathrm{E}_0}}{\frac{\sigma}{E_0}}\right)=\frac{F_1}{2}=\frac{10}{2}=5 \mathrm{~N}
\end{aligned}\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272292
A number of capacitors each of equal capacitance \(C\), are arranged as shown in Fig. The equivalent capacitance between \(A\) and \(B\) is
272293
A capacitor of capacity \(C_1\) is chargedupto \(V\) volt and then connected to an uncharged capacitor of capacity \(C_2\). Then final potential difference across each will be
1 \(\frac{C_2 V}{C_1+C_2}\)
2 \(\left(1+\frac{C_2}{C_1}\right) V\)
3 \(\frac{c_1 V}{c_1+c_2}\)
4 \(\left(1-\frac{c_2}{c_1}\right) V\)
Explanation:
(c) Common potential \(V=\frac{c_1 V+c_2 \times 0}{c_1+c_2}=\frac{c_1}{c_1+c_2} V\)
272290
Seven capacitors each of capacitance \(2 \mu F\) are to be connected in a configuration to obtain an effective capacitance of \(\left(\frac{10}{11}\right) \mu F\). Which of the combination \(\{5\}\) shown in figure will achieve the desired result?
1
2
3
4
Explanation:
(a) The equivalent capacitance
NCERT Page-78
Electrostatic Potentials and Capacitance
272291
A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be:
1 5 N
2 10 N
3 20 N
4 Zero
Explanation:
(a) Force on charge particle will be given by
\(\begin{aligned}
& F=q E \\
& F \propto E \\
& F_2=F_1\left(\frac{E_2}{E_1}\right)=F_1\left(\frac{\frac{\sigma}{2 \mathrm{E}_0}}{\frac{\sigma}{E_0}}\right)=\frac{F_1}{2}=\frac{10}{2}=5 \mathrm{~N}
\end{aligned}\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272292
A number of capacitors each of equal capacitance \(C\), are arranged as shown in Fig. The equivalent capacitance between \(A\) and \(B\) is
272293
A capacitor of capacity \(C_1\) is chargedupto \(V\) volt and then connected to an uncharged capacitor of capacity \(C_2\). Then final potential difference across each will be
1 \(\frac{C_2 V}{C_1+C_2}\)
2 \(\left(1+\frac{C_2}{C_1}\right) V\)
3 \(\frac{c_1 V}{c_1+c_2}\)
4 \(\left(1-\frac{c_2}{c_1}\right) V\)
Explanation:
(c) Common potential \(V=\frac{c_1 V+c_2 \times 0}{c_1+c_2}=\frac{c_1}{c_1+c_2} V\)
272290
Seven capacitors each of capacitance \(2 \mu F\) are to be connected in a configuration to obtain an effective capacitance of \(\left(\frac{10}{11}\right) \mu F\). Which of the combination \(\{5\}\) shown in figure will achieve the desired result?
1
2
3
4
Explanation:
(a) The equivalent capacitance
NCERT Page-78
Electrostatic Potentials and Capacitance
272291
A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be:
1 5 N
2 10 N
3 20 N
4 Zero
Explanation:
(a) Force on charge particle will be given by
\(\begin{aligned}
& F=q E \\
& F \propto E \\
& F_2=F_1\left(\frac{E_2}{E_1}\right)=F_1\left(\frac{\frac{\sigma}{2 \mathrm{E}_0}}{\frac{\sigma}{E_0}}\right)=\frac{F_1}{2}=\frac{10}{2}=5 \mathrm{~N}
\end{aligned}\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272292
A number of capacitors each of equal capacitance \(C\), are arranged as shown in Fig. The equivalent capacitance between \(A\) and \(B\) is
272293
A capacitor of capacity \(C_1\) is chargedupto \(V\) volt and then connected to an uncharged capacitor of capacity \(C_2\). Then final potential difference across each will be
1 \(\frac{C_2 V}{C_1+C_2}\)
2 \(\left(1+\frac{C_2}{C_1}\right) V\)
3 \(\frac{c_1 V}{c_1+c_2}\)
4 \(\left(1-\frac{c_2}{c_1}\right) V\)
Explanation:
(c) Common potential \(V=\frac{c_1 V+c_2 \times 0}{c_1+c_2}=\frac{c_1}{c_1+c_2} V\)