Electrostatic Potential and Capacitance NEET Questions Chapter-wise

0

072920 77839
info@smartachievers.in

Topic Question Set

Q 1 :
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The potential ( $V$ ) at any axial point, at 2 m distance ( $r$ ) from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4 \times 10^{- 6}$ C m, is $\pm 9 \times 10^{3}$ V.

Reason R: $V = \pm \frac{2 P}{4 π \in_{0} r^{2}},$ where $r$ is the distance of any axial point, situated at 2 m from the centre of the dipole.

In the light of the above statements, choose the correct answer from the options given below. [2024]

Both A and R are true and R is the correct explanation of A.
Both A and R are true and R is NOT the correct explanation of A.
A is true but R is false.
A is false but R is true.

1

2

3

4

(3)

Potential at any axial point of a dipole is

$V = \pm \frac{| \vec{p} |}{4 π ε_{0} r^{2}}$

$∴$ $V = \pm \frac{4 \times 10^{- 6} \times 9 \times 10^{9}}{{(2)}^{2}} = \pm 9 \times 10^{3}$ $V$

So, assertion is true but reason is false.

Q 2 :
A short electric dipole has a dipole moment of $16 \times 10^{- 9} C$ $m .$ The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of $60 °$ with the dipole axis is $(\frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} / C^{2})$ [2020]

50 V
200 V
400 V
zero

1

2

3

4

(2)

Potential due to dipole, $V = \frac{\vec{p} \cdot \hat{r}}{4 π ε_{0} r^{2}} = \frac{k p \cos θ}{r^{2}}$

or $V = \frac{(9 \times 10^{9}) (16 \times 10^{- 9}) \times \cos 60 °}{{(0.6)}^{2}} = 200$ $V$

Want Us to Email you About Special Offers & Updates?