There exists an electric field of magnitude E | vedclass

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

Question

(D) The work done $W$ by an electric field $E$ on a charge $q$ moving through a displacement $d$ at an angle $	heta$ with the field is given by the f

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) The work done $W$ by an electric field $E$ on a charge $q$ moving through a displacement $d$ at an angle $	heta$ with the field is given by the formula:$W = q E d \cos 	heta$Given:$q = 0.2 \, C$$d = 2 \, m$$	heta = 60^{\circ}$$W = 4 \, J$Substituting these values into the formula:$4 = (0.2) 	imes E 	imes 2 	imes \cos(60^{\circ})$Since $\cos(60^{\circ}) = 0.5$:$4 = 0.2 	imes E 	imes 2 	imes 0.5$$4 = 0.2 	imes E$$E = \frac{4}{0.2} = 20 \, N / C$Therefore,the value of $E$ is $20 \, N / C$.

There exists an electric field of magnitude $E$ in the $x$-direction. If the work done in moving a charge of $0.2 \, C$ through a distance of $2 \, m$ along a line making an angle $60^{\circ}$ with the $x$-axis is $4 \, J$,then the value of $E$ is ........ $N / C$.

Similar Questions

An observer at rest experiences an electric field due to a static charge. Later,he starts moving away from it. What does the observer experience?

Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is

In the figure shown,the distance from point $A$ to the point where the electric field intensity is zero is .......... $cm$.

$A$ cube of side $a$ has point charges $+Q$ located at each of its vertices except at the origin where the charge is $-Q$. The electric field at the centre of the cube is

What is the direction of electric field intensity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module