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 $\frac{q}{6 \pi \epsilon_0 a^2}$.
$(C)$ When $x=2q$,the potential at $O$ is $\frac{7q}{4 \sqrt{3} \pi \epsilon_0 a}$.
$(D)$ When $x=-3q$,the potential at $O$ is $\frac{3q}{4 \sqrt{3} \pi \epsilon_0 a}$.

  • A
    $A, B, C$
  • B
    $A, B, D$
  • C
    $A, B$
  • D
    $A, C$

Explore More

Similar Questions

Four point positive charges of same magnitude $(Q)$ are placed at four corners of a rigid square frame of side $L$ as shown in the figure. The plane of the frame is perpendicular to the $Z$-axis. If a negative point charge $(-q)$ is placed at a small distance $z$ away from the center of the frame along the $Z$-axis $(z < < L)$, then:

Four charges $q, 2q, -4q$ and $2q$ are placed in order at the four corners of a square of side $b$. The net field at the centre of the square is

Difficult
View Solution

Five point charges each having magnitude $q$ are placed at the corners of a regular hexagon as shown in the figure. The net electric field at the centre $O$ is $\vec E$. To make the net electric field at $O$ equal to $6\vec E$,the charge placed on the remaining sixth corner should be: (in $,q$)

Difficult
View Solution

Two point electric charges $+10^{-8} \text{ C}$ and $-10^{-8} \text{ C}$ are placed $0.1 \text{ m}$ apart. Find the magnitude of the total electric field at the centre of the line joining the two charges.

The charge distribution along the semi-circular arc is non-uniform. The charge per unit length $\lambda$ is given as $\lambda = \lambda_0 \sin \theta$,with $\theta$ measured as shown in the figure. $\lambda_0$ is a positive constant. The radius of the arc is $R$. The electric field at the center $P$ of the semi-circular arc is $E_1$. The value of $\frac{\lambda_0}{\epsilon_0 E_1 R}$ is

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo