Question

if the radius of a sphere is doubled what will be the electric field intensity on the new sphere

Answer 1

i think that it will be equal to the
$\frac{1} {4r}{2}$

Answer 2

Answer:

If the radius of a sphere is doubled and made a new sphere the electric field intensity on the new sphere is $E_{new} =E_{old} /4$

Explanation:

• The electric field due to a solid sphere with surface charge $q$ and radius $R$ at a distance $r$ from the center of the sphere (outside), is given by the Gauss law as

                        $E_{out} =\frac{q}{4\pi \varepsilon _0r^{2} }$

• The electric field on the surface of the solid sphere means $r=R$

                     $E_{on} =\frac{q}{4\pi \varepsilon _0R^{2} }$

          where the value of  $\frac{1}{4\pi \varepsilon _0} =9*10^{9}$

The electric field due to a sphere with surface charge $q$ and radius $R$ at the surface of the sphere is

$E_{on} =\frac{q}{4\pi \varepsilon _0R^{2} }$

Now the radius is doubled,

new radius is  $=2R$

electric field intensity of new sphere is $E_{new} =\frac{q}{4\pi \varepsilon _0(2R)^{2} }=\frac{1}{4} \frac{q}{4\pi \varepsilon _0R^{2} }$

that is $1/4$ times the old electric field

$E_{new} =E_{old} /4$