An electric dipole is kept on the axis of a u | vedclass

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

Question

(D) The electric field $E$ on the axis of a uniformly charged ring of radius $R$ and charge $Q$ at a distance $x$ from the center is given by $E = \fr

Vedclass · Accepted Answer

zero

Vedclass · Answer

(D) The electric field $E$ on the axis of a uniformly charged ring of radius $R$ and charge $Q$ at a distance $x$ from the center is given by $E = \frac{kQx}{(R^2 + x^2)^{3/2}}$.Given $x = R/\sqrt{2}$,the electric field is $E = \frac{kQ(R/\sqrt{2})}{(R^2 + R^2/2)^{3/2}} = \frac{kQR/\sqrt{2}}{(3R^2/2)^{3/2}} = \frac{kQR/\sqrt{2}}{3\sqrt{3}R^3/2\sqrt{2}} = \frac{2kQ}{3\sqrt{3}R^2}$.The force on a dipole $P$ in an electric field is $F = P \frac{dE}{dx}$.We need to find the derivative $\frac{dE}{dx}$ at $x = R/\sqrt{2}$.$\frac{dE}{dx} = kQ \left[ \frac{(R^2+x^2)^{3/2} - x \cdot \frac{3}{2}(R^2+x^2)^{1/2} \cdot 2x}{(R^2+x^2)^3} \right] = kQ \left[ \frac{R^2+x^2 - 3x^2}{(R^2+x^2)^{5/2}} \right] = kQ \frac{R^2-2x^2}{(R^2+x^2)^{5/2}}$.At $x = R/\sqrt{2}$,$R^2 - 2x^2 = R^2 - 2(R^2/2) = 0$.Therefore,the force $F = P \cdot 0 = 0$.

Similar Questions

The potential at a point $P$ due to an electric dipole is $1.8 \times 10^5 \, V$. If $P$ is at a distance of $50 \, cm$ from the centre $O$ of the dipole and the line $OP$ makes an angle of $60^{\circ}$ with the positive side of the axial line of the dipole,what is the dipole moment?

Choose the incorrect statement from the following.

$A$ system of three charges is placed as shown in the figure. If $D >> d$,the potential energy of the system is best given by:

Two point charges of same magnitude and opposite sign are fixed at points $A$ and $B$. $A$ third small point charge is to be balanced at point $P$ by the electrostatic force due to these two charges. The point $P$

Obtain the expression for the electric field due to an electric dipole at a point on its axial line.

Vedclass Test Series

Exam Paper Generator

Online Exam Module