Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance, Additional Exercise,Chapter-2 Exercise 2.21, NCERT Solutions for Class 12 Physics
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?
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Ans (a).
Zero at both the points
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= (1/4π ε0 ) [q/(z-a)] + (1/4π ε0 ) [-q/(z+a)]
= q (z+a -z + a )/ [4π ε0 (z²-a²)]
=2qa/[4π ε0 (z²-a²)]
=p/[4π ε0 (z²-a²)]
Where,
ε0 = Permittivity of free space and p = Dipole moment of the system of two charges = 2qa
Ans (b).
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 square of the distance,
i.e. V ∝ 1/r²
Ans (c).
Zero
The answer does not change if the path of the test is not along the x-axis.
A test charge is moved from point (5, 0, 0) to point (-7, 0, 0) along the x-axis.
Electrostatic potential (V₁) at point (5, 0, 0) is given by,
V₁= -(q /4π ε0) 1/[√(5-0)²+ (-a)²] + (q /4π ε0) 1/[√(5-0)²+ (a)²]
= -(q /4π ε0) 1/[√25+ a²] + (q /4π ε0) 1/[√25+ a²]
= 0
Electrostatic potential ,V₂ at point (-7,0,0) is given by
V₁= -(q /4π ε0) 1/[√(7-0)²+ (-a)²] + (q /4π ε0) 1/[√(7-0)²+ (a)²]
= -(q /4π ε0) 1/[√49+ a²] + (q /4π ε0) 1/[√49+ a²]
= 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.