JEE MAIN - Physics (2021 - 27th August Morning Shift - No. 1)
A uniformly charged disc of radius R having surface charge density $$\sigma$$ is placed in the xy plane with its center at the origin. Find the electric field intensity along the z-axis at a distance Z from origin :-
$$E = {\sigma \over {2{\varepsilon _0}}}\left( {1 - {Z \over {{{({Z^2} + {R^2})}^{1/2}}}}} \right)$$
$$E = {\sigma \over {2{\varepsilon _0}}}\left( {1 + {Z \over {{{({Z^2} + {R^2})}^{1/2}}}}} \right)$$
$$E = {{2{\varepsilon _0}} \over \sigma }\left( {{1 \over {{{({Z^2} + {R^2})}^{1/2}}}} + Z} \right)$$
$$E = {\sigma \over {2{\varepsilon _0}}}\left( {{1 \over {({Z^2} + {R^2})}} + {1 \over {{Z^2}}}} \right)$$
Explanation
Consider a small ring of radius r and thickness dr on disc.
_27th_August_Morning_Shift_en_1_1.png)
area of elemental ring on disc
dA = 2$$\pi$$rdr
charge on this ring dq = $$\sigma$$dA
$$dEz = {{kdqz} \over {{{({z^2} + {r^2})}^{3/2}}}}$$
$$E = \int\limits_0^R {d{E_z} = {\sigma \over {2{ \in _0}}}\left[ {1 - {z \over {\sqrt {{R^2} + {z^2}} }}} \right]} $$
area of elemental ring on disc
dA = 2$$\pi$$rdr
charge on this ring dq = $$\sigma$$dA
$$dEz = {{kdqz} \over {{{({z^2} + {r^2})}^{3/2}}}}$$
$$E = \int\limits_0^R {d{E_z} = {\sigma \over {2{ \in _0}}}\left[ {1 - {z \over {\sqrt {{R^2} + {z^2}} }}} \right]} $$
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