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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.
Asked by Aditi Last Modified
Answer 
Apoorva Purohit
There are two charges,
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.
Where,
= Permittivity of free space
For V = 0, equation (i) reduces to
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 s from the negative charge, where potential is zero, as shown in the following figure.
For this arrangement, potential is given by,
For V = 0, equation (ii) reduces to
Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.
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The potential at distance r cm from 5 × 10−8 C is zero. We know potential V=q/4π∈0r. So, V1+V2=0 5×10-8/4π∈0×r – 3×10-8/4π∈0(16-r) = 0 Solving we get r= 10 cm which is the zero possible point from 5 × 10−8 C
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Gourab Saha
Clear Your Concepts and Score Higher
Let, potential at distance r cm from 5 × 10−8 C is zero. We know potential V=q/4π∈0r. So,
V1+V2=0
5×10-8/4π∈0×r – 3×10-8/4π∈0(16-r) = 0
Solving we get r= 10 cm which is the zero potential point from 5 × 10−8 C
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CASE 1
There are two charges,
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.
Where,
= Permittivity of free space
For V = 0, equation (i) reduces to
Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges
CASE 2
Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential is zero,
as shown in the following figure.
For this arrangement, potential is given by,
For V = 0, equation (ii) reduces to
Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.
read less
V=kQ/r
Potential is a scalar. The points with zero potential will always be closer to the charge of smaller magnitude. There will be two points on the line joining the two charges, one in between the two charges and one on the extended line closer to the smaller charge.
kQ1/x + kQ2/(16-x)=0 will give 10 cm from positive charge.
kQ1/(16+x) + kQ2/(16-x)=0 will give 40 cm from positive charge.
Substitute charges with the sign.
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