Question

(i) If two similar large plates, each of area A having surface charge densities$+\sigma \text{and}-\sigma$ are separated by a distance in air, find the expressions for

(a) field at points between the two plates and on outer side of the plates. Specify the direction of the field in each case.

(b) the potential difference between the plates.

(c) the capacitance of the capacitor so formed.

(ii) Two metallic spheres of radii R and 2R are charged so that both of these have same surface charge density $\sigma$. If they are connected to each other with a conducting wire, in which direction will the charge flow and why ?

Answer 1

(i) (a) Consider a parallel plate capacitor with two identical plates X and Y, each having an area of A, and separated by a distance d. Let the space between the plates be filled by a dielectric medium with its dielectric constant as K and $\sigma$ be the surface charge density on each of the plates.

Surface charge density of plate 1
$\sigma =\frac{Q}{A}$
and that of plate 2 is $-\sigma$

Electric field in outer region I,
$E=\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}=0$

Electric field in outer region II,
$E=\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}=0$

In the inner region between plates 1 and 2,the electric fields due to the two charged plates add up.So

$E=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}$

(b) For uniform electric field,potential difference is simply the electric field multiplied by the distance between the plates,i.e.,

$V=Ed=\frac{1}{{\epsilon }_0}\frac{Qd}{A}$

(c) Now, the capacitance of the parallel plate
capacitor,

$C=\frac{Q}{V}=\frac{Q.{\epsilon }_{0}A}{Qd}=\frac{{\epsilon }_{0}A}{d}$


(ii) We know that the potential difference of the metallic sphere is given by,

$V=\frac{Q}{4\pi {\epsilon }_{0}r}$

where r is the radius of the sphere.

Now,the potential of the metallic sphere of radius R is given by,

$V_R=\frac{Q}{4\pi {\epsilon }_{0}r}$

$V_R=\frac{\sigma \left(4\pi R^2\right)}{4\pi {\epsilon }_{0}R}$

$V_R=\frac{\sigma R}{{\epsilon }_0}...\left(i\right)$

Similarly,potential of the metallic sphere of radius 2R is given by,

$V_{2}R=\frac{Q}{4\pi {\epsilon }_{0}2R}$

$V_{2}R=\frac{\sigma \left(4\pi \right(2R{)}^2)}{4\pi {\epsilon }_{0}2R}$

$V_{2}R=\frac{\sigma 2R}{{\epsilon }_0}...\left(ii\right)$

From the relation (i) and (ii) we know that $V_{2R}>V_R$. The charge will flow from the sphere of radius of 2R to the sphere of radius R, if the spheres are connected.