Question

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Q1.Two point charges 4Q, Q are separated by lm in air. At what point on the line joining the charges is the electric field intensity zero?
Also calculate the electrostatic potential energy of the system of charges, taking the value of charge, Q = 2 ×10^-7C.​​

Answer 1

Answer:

The point where, electric field intensity (E) is 0 on the line joining the charges.

(ii) Electrostatic potential energy (U) of the system of charges.

Let, the required point be P, at a distance x from charge 4Q and, at (1-x) distance from charge Q.

(i) Electric field at P due to charge 4Q = Electric field at P due to Q

E =kqr

Therefore, electric field intensity is 0 at x=23 m from charge 4Q.

(ii) Electrostatic potential energy of the system is given as,

Therefore, electrostatic potential energy of the ststem of charges is equal to 1.44 × 10-3 J .

Answer 2

Answer:

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