ABC is an equilateral triangle. Charges +q ar | vedclass

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

Question

(C) The electric field intensity due to a point charge $q$ at a distance $r$ is given by $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$.Since the tri

Vedclass · Accepted Answer

Zero

Vedclass · Answer

(C) The electric field intensity due to a point charge $q$ at a distance $r$ is given by $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$.Since the triangle $ABC$ is equilateral and the charges at each corner are identical $(+q)$,the distance from each corner to the centroid $O$ is the same $(r)$.Therefore,the magnitude of the electric field at $O$ due to each charge is $E_A = E_B = E_C = E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$.These three electric field vectors are directed away from the charges (outward from the corners) and are oriented at an angle of $120^\circ$ to each other.According to the principle of superposition,the net electric field at $O$ is the vector sum: $\vec{E}_{net} = \vec{E}_A + \vec{E}_B + \vec{E}_C$.Since the vectors are equal in magnitude and symmetrically distributed at $120^\circ$,their resultant is zero.Thus,the electric intensity at $O$ is zero.

$ABC$ is an equilateral triangle. Charges $+q$ are placed at each corner. The electric intensity at $O$ will be

Similar Questions

Find out the electric field intensity at point $A(1, 0, 2)$ due to a point charge $-20\,\mu C$ situated at point $B(0, 2, 1)$.

The unit $Volt/metre$ is the unit of:

Two point charges $Q_1$ and $Q_2$ are fixed at $x = 0$ and $x = a$. Assuming that the electric field strength is positive in the direction coinciding with the positive direction of $x$,which of the following options is correct?

Two infinitely long thin straight wires having uniform linear charge densities $2 \lambda$ and $3 \lambda$ are arranged parallel to each other at a distance $R$ apart. The intensity at a point midway between them is

In the given figure,the electric field at the center $O$ due to section $AB$ of a uniformly charged ring is $\overrightarrow{E}$. What will be the electric field at $O$ due to the remaining section $ACB$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module