JEE Advanced 2022 Paper 1, Physics Section 2 Q5: Coulomb's Law And Electric Field

JEE Advanced2022Paper 1Physics Section 2

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 $O$ of the hexagon and is bisected by the side.

Which of the following statement(s) is(are) correct in SI units?

A
When $x = q$ , the magnitude of the electric field at $O$ is zero.
B
When $x = - q$ , the magnitude of the electric field at $O$ is $\frac{q}{6 π ε _{0} a ^{2}}$ .
C
When $x = 2 q$ , the potential at $O$ is $\frac{7 q}{4 3 π ε _{0} a}$ .
D
When $x = - 3 q$ , the potential at $O$ is $- \frac{3 q}{4 3 π ε _{0} a}$ .

Solution

The distance from each external charge to centre $O$ is twice the apothem, i.e. $2 \cdot \frac{3 a}{2} = 3 a$ .

(A) Six identical charges arranged symmetrically about $O$ produce zero net field at $O$ . Correct.

(B) Replacing one $q$ with $- q$ is equivalent to the symmetric arrangement (zero field) plus an extra charge of $- 2 q$ at the position of $x$ . Field magnitude: $E = \frac{1}{4 π ε _{0}} \cdot \frac{2 q}{( 3 a ) ^{2}} = \frac{2 q}{4 π ε _{0} \cdot 3 a ^{2}} = \frac{q}{6 π ε _{0} a ^{2}}$ . Correct.

(C) Potentials add as scalars. With $x = 2 q$ , total charge contribution: $V = \frac{1}{4 π ε _{0} 3 a} (5 q + 2 q) = \frac{7 q}{4 3 π ε _{0} a}$ . Correct.

(D) With $x = - 3 q$ , $V = \frac{5 q + ( - 3 q )}{4 π ε _{0} 3 a} = \frac{2 q}{4 3 π ε _{0} a} = \frac{q}{2 3 π ε _{0} a}$ , not the value stated. Wrong.

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JEE Advanced 2022 Paper 1, Physics Section 2 Q5: Coulomb's Law And Electric Field

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