Physics Ncert Solutions Class 12th Physics Class 12th Electrostatic Potential and Capacitance

2.21 Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively.

(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?

(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.

(c) How much work is done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

3 Views | Posted 5 months ago

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Answered by

alok kumar singh | Contributor-Level 10

3 months ago

2.21 Charge –q is located at (0,0,-a) and charge +q is located at (0,0,a). Hence, they form a dipole. Point (0,0,z) is on the axis of this dipole and point (x,y,0) is normal to the axis of the dipole. Hence electrostatic potential at point (x,y,0) is zero. Electrostatic potential at point (0,0,z) is given by,

V = $\frac{1}{4 π ε_{0}}$ ( $\frac{q}{z - a}) + \frac{1}{4 π ε_{0}}$ ( $- \frac{q}{z + a})$ = $\frac{q (z + a - z + a)}{4 π ε_{0} (z^{2} - a^{2})}$ = $\frac{2 q a}{4 π ε_{0} (z^{2} - a^{2})}$ = $\frac{p}{4 π ε_{0} (z^{2} - a^{2})}$

Where $ε_{0}$ = Permittivity of free space

p = Dipole moment of the system of two charges = 2qa

Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to the dist

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Where $ε_{0}$ = Permittivity of free space

p = Dipole moment of the system of two charges = 2qa

Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to the distance. i.e. V $\propto$ $\frac{1}{r^{2}}$

Electrostatic potential ( $V_{1})$ at point (5,0,0) is given by

$V_{1}$ = $\frac{- q}{4 π ε_{0}} \frac{1}{\sqrt{{(5 - 0)}^{2} + {(- a)}^{2}}} + \frac{q}{4 π ε_{0}} \frac{1}{\sqrt{{(5 - 0)}^{2} + {(a)}^{2}}}$

= 0

Electrostatic potential ( $V_{2})$ at point (-7,0,0) is given by

$V_{2}$ = $\frac{- q}{4 π ε_{0}} \frac{1}{\sqrt{{(- 7)}^{2} + {(- a)}^{2}}} + \frac{q}{4 π ε_{0}} \frac{1}{\sqrt{{(- 7)}^{2} + {(a)}^{2}}}$

= 0

Hence, no work is done in moving a small test charge from point (5,0,0) to point (-7,0,0) along the x-axis.

The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

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