Question 828: Electrostatics & Capacitance

Question

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?

Accepted Answer

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 $$O = {1 \over {4\pi { \in _0}}}{{(2q)} \over {{{\left( {2 	imes {{\sqrt 3 a} \over 2}} ight)}^2}}}$$
$$ = {1 \over {4\pi { \in _0}}}{{2q} \over {3{a^2}}} = {q \over {6\pi { \in _0}{a^2}}}$$
$$\Rightarrow$$ (B) is correct.
For x = 2q, potential at O is
$${V_0} = 6 	imes {1 \over {4\pi { \in _0}}} 	imes {q \over {\sqrt 3 a}} + {1 \over {4\pi { \in _0}}}{q \over {\sqrt 3 a}}$$
$$ = {{7q} \over {4\sqrt 3 \pi { \in _0}a}}$$
$$\Rightarrow$$ (C) is correct.
For $$x =  - 3q,\,{V_0} = 2 	imes {1 \over {4\pi { \in _0}}} 	imes {q \over {\sqrt 3 a}} = {q \over {2\sqrt 3 \pi { \in _0}a}}$$
$$\Rightarrow$$ (D) is not correct.

Answer

When $x=q$, the magnitude of the electric field at 0 is zero.

Answer

When $x=-q$, the magnitude of the electric field at 0 is $\frac{q}{6 \pi \epsilon_{o} a^{2}}$.

Answer

When $x=2 q$, the potential at 0 is $\frac{7 q}{4 \sqrt{3} \pi \epsilon_{o} a}$.

Answer

When $x=-3 q$, the potential at 0 is $-\frac{3 q}{4 \sqrt{3} \pi \epsilon_{o} a}$.

Electrostatics & Capacitance

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