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? (b) Obtain the dependence of - Physics

Numerical

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?

(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?

Solution

(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/(4piin_0)("q"/("z" - "a")) + 1/(4piin_0)(-"q"/("z" + "a"))

= ("q"("z" + "a" - "z" + "a"))/(4piin_0("z"^2 - "a"^2))

= (2"qa")/(4piin_0("z"^2 - "a"^2))

= "p"/(4piin_0("z"^2 - "a"^2))

Where,

in_0 = Permittivity of free space

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

(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 the square of the distance i.e., "V" prop 1/"r"^2

(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,
"V"_1 = (-"q")/(4piin_0)1/sqrt((5-0)^2 + (-"a")^2) + "q"/(4piin_0)1/(sqrt((5 - 0)^2)+("a")^2)

= (-"q")/(4piin_0sqrt(25 + "a"^2)) + "q"/(4piin_0sqrt(25 + "a"^2))

= 0

Electrostatic potential, V2, at point (− 7, 0, 0) is given by,

"V"_2 = (-"q")/(4piin_0) 1/sqrt((-7)^2 + (-"a")^2) + "q"/(4piin_0) 1/sqrt((-7)^2 + ("a")^2)

= (-"q")/(4piin_0sqrt((49 + "a"^2))) + "q"/(4piin_0sqrt((49 + "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.

Concept: Potential Energy in an External Field - Potential Energy of a System of Two Charges in an External Field
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APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.21 | Page 88
NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 21 | Page 90
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