Learn steps to capacitance of a capacitor calculation, practice to calculation of potential difference between charged shells and spherical & cylindrical capacitor.
A Charge (Q)\(\rightarrow\) Electric field (E)\(\rightarrow\) Potential difference(\(\Delta V\)) \(\rightarrow\)Capacitance, \(C=\dfrac {Q}{\Delta V}\)
B Charge (Q)\(\rightarrow\) Potential difference(\(\Delta V\)) \(\rightarrow\) Electric field (E)\(\rightarrow\)Capacitance, \(C=\dfrac {\Delta V}{E}\)
C Charge (Q)\(\rightarrow\) Potential difference(\(\Delta V\)) \(\rightarrow\) Electric field (E)\(\rightarrow\)Capacitance, \(C=\dfrac {Q}{E}\)
D None of these
Step 1: The electric field at a distance 'r' from a line charge of infinite length is given as \(E=\dfrac {\lambda }{2\pi\epsilon_0r}\)
where \(\lambda\) = charge per unit length
Step 2 : Potential difference in terms of electric field is given as
\(\Delta V = -\int\limits_a^b\;\vec E\cdot\vec ds\)
\(V_b-V_a= -\int\limits_a^b\;\vec E\cdot\vec ds\)
\(V_b-V_a=-\int\limits_a^b \dfrac {\lambda}{2\pi\epsilon_0r}dr\)
\(\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0}\int\limits_a^b\dfrac {1}{r} dr\)
\(\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0} [\ell n\;r]_a^b\)
\(\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0} [\ell n\,b-\ell n\,a]\)
\(\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0} \left [\ell n(\dfrac {b}{a})\right]\)
\(\left [\lambda=\dfrac {Q}{\ell}\text{where }\lambda \,\,\text{is charge per unit length} \right]\)
\(\Rightarrow V_b-V_a=-\dfrac {Q}{2\pi\epsilon_0\ell}\ell n \left( \dfrac {b}{a} \right)\)
The magnitude of the potential difference is
\( \Delta V=|V_b-V_a|=\dfrac {Q}{2\pi\epsilon_0\ell}\ell n \left( \dfrac {b}{a} \right)\)
Step 3 : Capacitance of cylindrical capacitor is given as
\(C=\dfrac {Q}{\Delta V}=\dfrac {Q}{V_b-V_a}\)
\(\Rightarrow C=\dfrac {Q} {\dfrac {Q}{2\pi\epsilon_0 \ell}\ell n\left (\dfrac {b}{a}\right)}\)
\(\Rightarrow C=\dfrac {\ell}{2K\ell n\left (\dfrac {b}{a}\right)}\)
A \(\dfrac{1}{\ell n\,(4)} nF\)
B \(\dfrac{2}{\ell n\,(5)} nF\)
C \(\dfrac{1}{\ell n\,(2)} nF\)
D \(\dfrac{2}{\ell n\,(2)} nF\)
A \(\dfrac {2}{1}\)
B \(\dfrac {3}{1}\)
C \(\dfrac {4}{3}\)
D \(\dfrac {6}{5}\)
Step 1 : Consider a spherical conducting shell of radius b, having charge –Q concentric with a smaller conducting sphere of radius a, having charge +Q.
Step 2: Electric field outside the sphere is in radial direction (if radius is r)
\(E=\dfrac {KQ}{r^2}\)
Step 3 : Potential difference between a conducting sphere and a conducting shell, is given as
\(V_b-V_a=-\int\limits_a^b\vec E\cdot\vec dr\)
or, \(V_b-V_a=-\int\limits_a^b\dfrac {KQ}{r^2}\;dr\)
or, \(V_b-V_a=-KQ\int\limits_a^b\dfrac {1}{r^2}\;dr\)
or, \(V_b-V_a=-KQ \left [ -\dfrac {1}{r} \right]_a^b\)
or, \(V_b-V_a=KQ \left [ \dfrac {1}{b}-\dfrac {1}{a} \right]\)
The magnitude of the potential difference is
\(\Delta V \,=\,\, |V_b-V_a|=\dfrac {KQ(b-a)}{ab} \)
Step 4 : Capacitance of the body is given as
\(C=\dfrac {Q}{\Delta V}\)
or, \(C=\dfrac {Q}{KQ\left ( \dfrac {b-a}{ab}\right)}\)
or, \(C=\dfrac {ab}{K(b-a)}\)
A Charge (Q)\(\rightarrow\) Potential difference(\(\Delta V\)) \(\rightarrow\)Capacitance, \(C=\dfrac {\Delta V}{Q}\)
B Charge (Q)\(\rightarrow\)Electric field (E)\(\rightarrow\) Potential difference(\(\Delta V\)) \(\rightarrow\)Capacitance, \(C=\dfrac {Q}{\Delta V}\)
C Charge (Q)\(\rightarrow\)Potential difference(\(\Delta V\)) \(\rightarrow\)Electric field (E)\(\rightarrow\) Capacitance, \(C=\dfrac {Q}{\Delta V}\)
D None of these
Step 1 : Consider a spherical conducting shell of radius b, having charge –Q concentric with a smaller conducting sphere of radius a, having charge +Q.
Step 2: Electric field outside the sphere is in radial direction (if radius is r)
\(E=\dfrac {KQ}{r^2}\)
Step 3 : Potential difference between a conducting sphere and a conducting shell, is given as
\(V_b-V_a=-\int\limits_a^b\vec E\cdot\vec dr\)
or, \(V_b-V_a=-\int\limits_a^b\dfrac {KQ}{r^2}\;dr\)
or, \(V_b-V_a=-KQ\int\limits_a^b\dfrac {1}{r^2}\;dr\)
or, \(V_b-V_a=-KQ \left [- \dfrac {1}{r} \right]_a^b\)
or, \(V_b-V_a=KQ \left [ \dfrac {1}{b}-\dfrac {1}{a} \right]\)
The magnitude of the potential difference is
or, \(\Delta V=\,\,|V_b-V_a|=\dfrac {KQ(b-a)}{ab} \)
Step 4 : Capacitance of the body is given as
\(C=\dfrac {Q}{\Delta V}\)
or, \(C=\dfrac {Q}{KQ\left ( \dfrac {b-a}{ab}\right)}\)
or, \(C=\dfrac {ab}{K(b-a)}\)
The electric field lines for the inner conductor will be radially outward.
means \(b\rightarrow\infty\)
\(C= \displaystyle\lim_{b\to\infty} \dfrac {ab}{K(b-a)}\)
or, \(C=\displaystyle\lim_{b\to\infty}\dfrac {a}{K\left ( 1-\dfrac {a}{b}\right)}\) \(\left [ \displaystyle\lim_{b\to\infty}\dfrac {a}{b}=0 \right]\)
\(C=\dfrac {a}{K}\)
\(C=4\pi\epsilon_0a\)
This is the capacitance for isolated sphere.
A \(20\pi\epsilon_0\,F\)
B \(5\pi\epsilon_0\,F\)
C \(10\pi\epsilon_0\,F\)
D \(\pi\epsilon_0\,F\)