Four equal charges of value +Q are placed at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In a regular hexagon, the distance from the centre $O$ to any vertex is equal to the side length $a$. To maximize the electric field at the centre

Vedclass · Accepted Answer

$\frac{\sqrt{3}Q}{4 \pi \epsilon _{0} a^{2}}$

Vedclass · Answer

(C) In a regular hexagon, the distance from the centre $O$ to any vertex is equal to the side length $a$. To maximize the electric field at the centre, we place the four charges at vertices $A, B, C,$ and $F$ as shown in the figure. The electric field due to charges at $F$ and $C$ are equal in magnitude and opposite in direction, so they cancel each other out $(E_F + E_C = 0)$. The net electric field at the centre $O$ is due to the charges at $A$ and $B$. The magnitude of the electric field due to each charge is $E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{a^2}$. The angle between the electric field vectors $E_A$ and $E_B$ is $60^{\circ}$. The resultant electric field $E_{	ext{net}}$ is given by: $E_{	ext{net}} = \sqrt{E_A^2 + E_B^2 + 2 E_A E_B \cos 60^{\circ}}$ Since $E_A = E_B = E$, we have: $E_{	ext{net}} = \sqrt{E^2 + E^2 + 2 E^2 (1/2)} = \sqrt{3E^2} = E\sqrt{3}$ Substituting the value of $E$: $E_{	ext{net}} = \frac{\sqrt{3}Q}{4 \pi \epsilon_0 a^2}$

Four equal charges of value $+Q$ are placed at four vertices of a regular hexagon of side $a$. By suitably choosing the vertices, what can be the maximum possible magnitude of the electric field at the centre of the hexagon?

Similar Questions

Two charges,each equal to $\eta q$ (where $\eta^{-1} < \sqrt{3}$),are placed at two corners of an equilateral triangle of side $a$. The electric field at the third corner is $E_3$. If $E_0 = \frac{q}{4\pi\varepsilon_0 a^2}$,then which of the following is correct?

Four point charges $-q, +q, +q$ and $-q$ are placed on the $y$-axis at $y = -2d, y = -d, y = +d$ and $y = +2d$,respectively. The magnitude of the electric field $E$ at a point on the $x$-axis at $x = D$,with $D >> d$,will vary as:

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.

Vedclass Test Series

Exam Paper Generator

Online Exam Module