CBSE Class 12 Physics 2020 Delhi Set 1 Paper

Question : 33

Total: 37

(a) Two point charges q1 and q2 are kept at a distance of r12 in air. Deduce the expression for the electrostatic potential energy of the system.
(b) If an external electric field (E) is applied on the system, write the expression for total energy of this system.

Solution:

(a)

When q1 is placed, potential at r2=V2=k‌

r12

Potential energy when q2 is placed at r2
=U1=q2V2=kq2‌

r12

When q2 is placed, potential at r1=V1=k‌

r12

Potential energy when q1 is placed at r1
=U2=q1V1=kq1‌

r12

Potential energy of the system
=1∕2[kq2‌

r12

+kq1‌

r12

] =k‌

q1q2

r12

(b) An external field E is applied on the system : q1 and q2 are two charges located at r1 and r2. in an external electric field (E).
Work done in bringing q1 from infinity to r1=W1= q1Vr1
Work done on q2 for bringing it from infinity to r2 against the external field (E)=W2=q2Vr2
Work done on q2 against the field due to q1=W12=‌

q1q2

4πε0r12

(r12= distance between q1 and q2)
So, the potential energy of the system =W1+W2+ W12
U=q1Vr1+q2Vr2+‌

q1q2

4πε0r12

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