JEE Advance - Physics (2022 - Paper 1 Online - No. 13)
Six charges are placed around a regular hexagon of side length $a$ as shown in the figure. Five of them have charge $q$, and the remaining one has charge $x$. The perpendicular from each charge to the nearest hexagon side passes through the center 0 of the hexagon and is bisected by the side.
Which of the following statement(s) is(are) correct in SI units?
Explanation
When, x = q, the situation is symmetric
$$\Rightarrow$$ Electric field at O would be zero.
$$\Rightarrow$$ (A) is correct.
When x = $$-$$q, we can think of x as q + ($$-$$2q) $$\Rightarrow$$ Magnitude of electric field at $$O = {1 \over {4\pi { \in _0}}}{{(2q)} \over {{{\left( {2 \times {{\sqrt 3 a} \over 2}} \right)}^2}}}$$
$$ = {1 \over {4\pi { \in _0}}}{{2q} \over {3{a^2}}} = {q \over {6\pi { \in _0}{a^2}}}$$
$$\Rightarrow$$ (B) is correct.
For x = 2q, potential at O is
$${V_0} = 6 \times {1 \over {4\pi { \in _0}}} \times {q \over {\sqrt 3 a}} + {1 \over {4\pi { \in _0}}}{q \over {\sqrt 3 a}}$$
$$ = {{7q} \over {4\sqrt 3 \pi { \in _0}a}}$$
$$\Rightarrow$$ (C) is correct.
For $$x = - 3q,\,{V_0} = 2 \times {1 \over {4\pi { \in _0}}} \times {q \over {\sqrt 3 a}} = {q \over {2\sqrt 3 \pi { \in _0}a}}$$
$$\Rightarrow$$ (D) is not correct.
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