272282
Three capacitors are connected in the arms of a triangle \(A B C\) as shown in figure 5 V is applied between \(A\) and \(B\). The voltage between\(B\) and \(C\) is
1 2 V
2 \(3 V\)
3 1 V
4 1.5 V
Explanation:
(a) series \(=1 \mu \mathrm{~F}\)
\(\frac{1}{\mathrm{C}_{\text {series }}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\frac{3+2+1}{6}=\frac{6}{6}\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272283
To obtain \(3 \mu \mathrm{~F}\) capacity from three capacitors of \(2 \mu \mathrm{~F}\) each, they will be arranged.
1 all the three in series
2 all the three in parallel
3 two capacitors in series and the third in parallel with the combinatioin of first two
4 two capacitors in parallel and the third in series with the combination of first two
Explanation:
\(\left\{c\right.\) c) \(C=\frac{2 \times 2}{2+2}+2=3 \mu F\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272284
The energy required to charge a parallel plate condenser of plate separation \(d\) and plate area of cross-section \(A\) such that the uniform electric field between the plates is \(E\), is
(a) Energy required to charge the capacitor is \(W=U=Q V\)
\(\begin{array}{r}
\Rightarrow \quad U^{\prime}=C V^2=\frac{\varepsilon_0 A}{d} \cdot V^2=\frac{\varepsilon_0 A d}{d^2} \cdot V^2=\varepsilon_0 E^2 A d \\
{\left[\because E=\frac{V}{d}\right] .}
\end{array}\)
NCERT Page-79/N-74
Electrostatic Potentials and Capacitance
272285
Two identical thin metal plates has charge \(q_1\) and \(q_2\)
respectively such that \(q_1>q_2\). The plates were brought close to each other to form a parallel plate capacitor of capacitance \(C\). The potential difference between them is:
1 \(\frac{\left(q_1+q_2\right\}}{c}\)
2 \(\frac{\left(q_1-q_2\right)}{c}\)
3 \(\frac{\left(q_1-q_2\right)}{2 C}\)
4 \(\frac{2\left(q_1-q_2\right)}{c}\)
Explanation:
(c) Electric field between plates given by, \(E=\frac{q_1-q_2}{2 A E_0}\)
\{Here, \(\mathrm{q}_1>\mathrm{q}_2\) \}
The, the potential difference will be
\(\begin{aligned}
v=E d=\frac{q_1-q_2}{2 A E_0} d=\frac{q_1-q_2}{2 C} & \\
& \left(\because C=\frac{\epsilon_0 A}{d}\right)
\end{aligned}\)
272282
Three capacitors are connected in the arms of a triangle \(A B C\) as shown in figure 5 V is applied between \(A\) and \(B\). The voltage between\(B\) and \(C\) is
1 2 V
2 \(3 V\)
3 1 V
4 1.5 V
Explanation:
(a) series \(=1 \mu \mathrm{~F}\)
\(\frac{1}{\mathrm{C}_{\text {series }}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\frac{3+2+1}{6}=\frac{6}{6}\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272283
To obtain \(3 \mu \mathrm{~F}\) capacity from three capacitors of \(2 \mu \mathrm{~F}\) each, they will be arranged.
1 all the three in series
2 all the three in parallel
3 two capacitors in series and the third in parallel with the combinatioin of first two
4 two capacitors in parallel and the third in series with the combination of first two
Explanation:
\(\left\{c\right.\) c) \(C=\frac{2 \times 2}{2+2}+2=3 \mu F\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272284
The energy required to charge a parallel plate condenser of plate separation \(d\) and plate area of cross-section \(A\) such that the uniform electric field between the plates is \(E\), is
(a) Energy required to charge the capacitor is \(W=U=Q V\)
\(\begin{array}{r}
\Rightarrow \quad U^{\prime}=C V^2=\frac{\varepsilon_0 A}{d} \cdot V^2=\frac{\varepsilon_0 A d}{d^2} \cdot V^2=\varepsilon_0 E^2 A d \\
{\left[\because E=\frac{V}{d}\right] .}
\end{array}\)
NCERT Page-79/N-74
Electrostatic Potentials and Capacitance
272285
Two identical thin metal plates has charge \(q_1\) and \(q_2\)
respectively such that \(q_1>q_2\). The plates were brought close to each other to form a parallel plate capacitor of capacitance \(C\). The potential difference between them is:
1 \(\frac{\left(q_1+q_2\right\}}{c}\)
2 \(\frac{\left(q_1-q_2\right)}{c}\)
3 \(\frac{\left(q_1-q_2\right)}{2 C}\)
4 \(\frac{2\left(q_1-q_2\right)}{c}\)
Explanation:
(c) Electric field between plates given by, \(E=\frac{q_1-q_2}{2 A E_0}\)
\{Here, \(\mathrm{q}_1>\mathrm{q}_2\) \}
The, the potential difference will be
\(\begin{aligned}
v=E d=\frac{q_1-q_2}{2 A E_0} d=\frac{q_1-q_2}{2 C} & \\
& \left(\because C=\frac{\epsilon_0 A}{d}\right)
\end{aligned}\)
272282
Three capacitors are connected in the arms of a triangle \(A B C\) as shown in figure 5 V is applied between \(A\) and \(B\). The voltage between\(B\) and \(C\) is
1 2 V
2 \(3 V\)
3 1 V
4 1.5 V
Explanation:
(a) series \(=1 \mu \mathrm{~F}\)
\(\frac{1}{\mathrm{C}_{\text {series }}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\frac{3+2+1}{6}=\frac{6}{6}\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272283
To obtain \(3 \mu \mathrm{~F}\) capacity from three capacitors of \(2 \mu \mathrm{~F}\) each, they will be arranged.
1 all the three in series
2 all the three in parallel
3 two capacitors in series and the third in parallel with the combinatioin of first two
4 two capacitors in parallel and the third in series with the combination of first two
Explanation:
\(\left\{c\right.\) c) \(C=\frac{2 \times 2}{2+2}+2=3 \mu F\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272284
The energy required to charge a parallel plate condenser of plate separation \(d\) and plate area of cross-section \(A\) such that the uniform electric field between the plates is \(E\), is
(a) Energy required to charge the capacitor is \(W=U=Q V\)
\(\begin{array}{r}
\Rightarrow \quad U^{\prime}=C V^2=\frac{\varepsilon_0 A}{d} \cdot V^2=\frac{\varepsilon_0 A d}{d^2} \cdot V^2=\varepsilon_0 E^2 A d \\
{\left[\because E=\frac{V}{d}\right] .}
\end{array}\)
NCERT Page-79/N-74
Electrostatic Potentials and Capacitance
272285
Two identical thin metal plates has charge \(q_1\) and \(q_2\)
respectively such that \(q_1>q_2\). The plates were brought close to each other to form a parallel plate capacitor of capacitance \(C\). The potential difference between them is:
1 \(\frac{\left(q_1+q_2\right\}}{c}\)
2 \(\frac{\left(q_1-q_2\right)}{c}\)
3 \(\frac{\left(q_1-q_2\right)}{2 C}\)
4 \(\frac{2\left(q_1-q_2\right)}{c}\)
Explanation:
(c) Electric field between plates given by, \(E=\frac{q_1-q_2}{2 A E_0}\)
\{Here, \(\mathrm{q}_1>\mathrm{q}_2\) \}
The, the potential difference will be
\(\begin{aligned}
v=E d=\frac{q_1-q_2}{2 A E_0} d=\frac{q_1-q_2}{2 C} & \\
& \left(\because C=\frac{\epsilon_0 A}{d}\right)
\end{aligned}\)
272282
Three capacitors are connected in the arms of a triangle \(A B C\) as shown in figure 5 V is applied between \(A\) and \(B\). The voltage between\(B\) and \(C\) is
1 2 V
2 \(3 V\)
3 1 V
4 1.5 V
Explanation:
(a) series \(=1 \mu \mathrm{~F}\)
\(\frac{1}{\mathrm{C}_{\text {series }}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\frac{3+2+1}{6}=\frac{6}{6}\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272283
To obtain \(3 \mu \mathrm{~F}\) capacity from three capacitors of \(2 \mu \mathrm{~F}\) each, they will be arranged.
1 all the three in series
2 all the three in parallel
3 two capacitors in series and the third in parallel with the combinatioin of first two
4 two capacitors in parallel and the third in series with the combination of first two
Explanation:
\(\left\{c\right.\) c) \(C=\frac{2 \times 2}{2+2}+2=3 \mu F\)
NCERT Page-78
Electrostatic Potentials and Capacitance
272284
The energy required to charge a parallel plate condenser of plate separation \(d\) and plate area of cross-section \(A\) such that the uniform electric field between the plates is \(E\), is
(a) Energy required to charge the capacitor is \(W=U=Q V\)
\(\begin{array}{r}
\Rightarrow \quad U^{\prime}=C V^2=\frac{\varepsilon_0 A}{d} \cdot V^2=\frac{\varepsilon_0 A d}{d^2} \cdot V^2=\varepsilon_0 E^2 A d \\
{\left[\because E=\frac{V}{d}\right] .}
\end{array}\)
NCERT Page-79/N-74
Electrostatic Potentials and Capacitance
272285
Two identical thin metal plates has charge \(q_1\) and \(q_2\)
respectively such that \(q_1>q_2\). The plates were brought close to each other to form a parallel plate capacitor of capacitance \(C\). The potential difference between them is:
1 \(\frac{\left(q_1+q_2\right\}}{c}\)
2 \(\frac{\left(q_1-q_2\right)}{c}\)
3 \(\frac{\left(q_1-q_2\right)}{2 C}\)
4 \(\frac{2\left(q_1-q_2\right)}{c}\)
Explanation:
(c) Electric field between plates given by, \(E=\frac{q_1-q_2}{2 A E_0}\)
\{Here, \(\mathrm{q}_1>\mathrm{q}_2\) \}
The, the potential difference will be
\(\begin{aligned}
v=E d=\frac{q_1-q_2}{2 A E_0} d=\frac{q_1-q_2}{2 C} & \\
& \left(\because C=\frac{\epsilon_0 A}{d}\right)
\end{aligned}\)
