Question 225: Electrostats & Capacitance

Question

Let $${E_1}\left( r \right),{E_2}\left( r \right)$$ and $${E_3}\left( r \right)$$ be the respective electric field at a distance $$r$$ from a point charge $$Q,$$ an infinitely long wire with constant linear charge density $$\lambda ,$$ and an infinite plane with uniform surface charge density $$\sigma .$$ If $$E{}_1\left( {{r_0}} \right) = {E_2}\left( {{r_0}} \right) = {E_3}\left( {{r_0}} \right)$$ at  a given distance $${r_0}.$$ then

Accepted Answer

$${E_1}(r) = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{r^2}}}$$
$${E_2}(r) = {\lambda  \over {2\pi {\varepsilon _0}r}}$$
$${E_3}(r) = {\sigma  \over {2{\varepsilon _0}}}$$
At $$r = {r_0}$$,
$${E_1}({r_0}) = {1 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}}$$
$${E_2}({r_0}) = {\lambda  \over {2\pi {\varepsilon _0}{r_0}}}$$
$${E_3}({r_0}) = {\sigma  \over {2{\varepsilon _0}}}$$
As $${E_1}({r_0}) = {E_2}({r_0}) = {E_3}({r_0})$$ (Given)
$$	herefore$$ $${1 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}} = {\lambda  \over {2\pi {\varepsilon _0}{r_0}}} = {\sigma  \over {2{\varepsilon _0}}}$$
Then
$${1 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}} = {\sigma  \over {2{\varepsilon _0}}}$$
or $$Q = \sigma \pi r_0^2$$
Hence, option (a) is incorrect.
Now, $${\lambda  \over {2\pi {\varepsilon _0}{r_0}}} = {\sigma  \over {2{\varepsilon _0}}}$$ or $${r_0} = {\lambda  \over {\pi \sigma }}$$
Hence, option (b) is incorrect.
At $$r = {r_0}/2$$,
$${E_1}({r_0}/2) = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{{({r_0}/2)}^2}}} = {4 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}} = 4{E_1}({r_0})$$
or, $${E_1}({r_0}) = {1 \over 4}{E_1}({r_0}/2)$$
$${E_2}({r_0}/2) = {\lambda  \over {2\pi {\varepsilon _0}({r_0}/2)}} = {{2\lambda } \over {2\pi {\varepsilon _0}{r_0}}} = 2{E_2}({r_0})$$
or, $${E_2}({r_0}) = {1 \over 2}{E_2}({r_0}/2)$$
$$\because $$$${E_1}({r_0}) = {E_2}({r_0})$$
$$	herefore$$ $${1 \over 4}{E_1}({r_0}/2) = {1 \over 2}{E_2}({r_0}/2)$$
or, $${E_1}({r_0}/2) = 2{E_2}({r_0}/2)$$
Hence, option (c) is correct.
$${E_3}({r_0}/2) = {\sigma  \over {2{\varepsilon _0}}} = {E_3}({r_0})$$
$$\because$$ $${E_2}({r_0}) = {E_3}({r_0})$$
$$	herefore$$ $${1 \over 2}{E_2}({{{r_0}} \over 2}) = {E_3}({{{r_0}} \over 2})$$
or, $${E_2}({r_0}/2) = 2{E_3}({r_0}/2)$$
Hence, option (d) is incorrect.

Answer

$$Q = 4\sigma \pi r_0^2$$

Answer

$${r_0} = {\lambda  \over {2\pi \sigma }}$$

Answer

$${E_1}\left( {{r_0}/2} \right) = 2{E_2}\left( {{r_0}/2} \right)$$

Answer

$${E_2}\left( {{r_0}/2} \right) = 4{E_3}\left( {{r_0}/2} \right)$$

Electrostats & Capacitance

Choose the correct answer: