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Two Charges 5 × 10^−8 C and −3 × 10^−8 C Are Located 16 cm Apart. at What Point(S) on the Line Joining the Two Charges is the Electric Potential Zero? Take the Potential at Infinity to Be Zero. - Physics

Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

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Solution

There are two charges,

`q_1=5xx10^-8C`

`q_2=-3xx10^-8C`

Distance between the two charges, d = 16 cm = 0.16 m

Consider a point P on the line joining the two charges, as shown in the given figure.

r = Distance of point P from charge q1

Let the electric potential (V) at point P be zero.

Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.

`therefore V=q_1/(4piin_0r)+q_2/(4piin_0(d-r))`

Where,

`in_0`= Permittivity of free space

For V = 0, equation (i) reduces to

`q_1/(4piin_0r)=-q_2/(4piin_0(d-r))`

`q_1/r=-q_2/(d-r)`

`(5xx10^-8)/r=-(-3xx10^-8)/(0.16-r)`

`0.16/r=8/5`

`therefor r=0.1m=10cm`

Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.

Suppose point P is outside the system of two charges at a distance from the negative charge, where potential is zero, as shown in the following figure.

For this arrangement, potential is given by,

`V=q_1/(4piin_0s)+q_2/(4 piin_0(s-d))`

For V = 0, equation (ii) reduces to

`q_1/(4piin_0s)=q_2/(4piin_0(s-d))`

`q_1/s=-q_2/(s-d)`

`(5xx10^-8)/s=(-3xx10^-8)/(s-0.16)`

`1-0.16/s=3/5`

`0.16/s=2/5`

`therefore s=0.4m=40 cm`

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

Concept: Potential Due to an Electric Dipole
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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Q 1 | Page 87
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