Q. 35

Two particles A and B having charges of +2.00 × 10–6 C and of –4.00 × 10–6 C respectively are held fixed at a separation of 20.0 cm. Locate the point(s) on the line AB where (a) electric field is zero (b) the electric potential is zero.

Answer :


Given:
Charge on particle A: qA = +2× 10-6 C
Charge on particle B : qB = -4× 10-6 C
Separation between A and B : r = 20.0 cm=0.2m

Formula used:
Electric Field given as:
Here, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is the distance between two charges.
Also,
Electric potential is given as:
Here, k is a constant and k= =9× 109 Nm2C-2 .q is the point charge and r is the distance between two charges.
Now,

Let there be a point at distance ‘x’ from A where net electric field is zero.
Distance between A and the point be : rA = x m
Distance between B and the point be : rB = (0.2-x) m










Hence, electric field is zero at a point 0.4825 m from A along AB.
Now,
Zero Net potential at the point is








OR

Hence, potential can be zero at 0.0666 m from A along AB and at 0.2 m from B along AB.


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