Question

A uniformly charged spherical conductor has a radius of 1m and surface charge density $0.8\frac{C}{m^2}$. The electric field magnitude because of this at a point 1m radially away, from the surface of the sphere is

• A. $4.5\times {10}^9\frac{N}{C}$
• B. $9\times {10}^9\frac{N}{C}$
• C. $2.25\times {10}^{10}\frac{N}{C}$
• D. $7.2\times {10}^9\frac{N}{C}$

Answer 1

The correct option is C

$2.25\times {10}^{10}\frac{N}{C}$

While calculating electric field at any outside point (outside the surface I mean), a sphere behaves like a point charge and the electric field magnitude is given by

Here it is given $\sigma =0.8\frac{C}{m^2}$

As $Q=\sigma \times area$

$=\sigma \times 4\pi R^2$

We have $Q=0.8\times 4\pi \times 1^2$

$\approx 10C$

Remember that the ‘r’ in the equation of electric field is the distance from the centre of the sphere and not from its surface.

So here r = R + distance from surface

= 1m + 1m (given)

=2m

$\therefore |\overrightarrow{E}|=\frac{9\times {10}^9\times 10C}{2^2}=2.25\times {10}^{10}\frac{N}{C}$