CBSE Class 12 Physics 2018 Delhi Set 1 Paper

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Question : 11
Total: 26
SECTION -C

Four point charges Q,q,Q and q are placed at the corners of a square of side ' a ' as shown in the figure.

Find the
(a) resultant electric force on a charge Q and
(b) potential energy of this system.
OR
(a) Three point charges q,−4q and 2q are placed at the vertices of an equilateral triangle ABC of side ' l ' as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.

(b) Find out the amount of the work done to separate the charges at infinite distance.
Solution:  
(a) Finding the resultant force on a charge Q
(b) Potential Energy of the system
(a) Let us find the force on the charge Q at the point C
F1=‌
1
4πε0
‌
Q2
(a√2)2
=‌
1
4πε0
(‌
Q2
2a2
)
‌ (along ‌AC‌ )
Force due to the charge q (at B ),
F2=‌
1
4πε0
‌
qQ
a2
‌ along
‌BC

Force due to the charge q (at D ),
F3=‌
1
4πε0
‌
qQ
a2
‌ along
‌DC

Resultant of these two equal forces

Force due to the other charge Q
F23=‌
1
4πε0
‌
qQ(√2)
a2
‌ (along
‌AC
‌ )
‌

∴ Net force on charge Q ( at point C )
F=F1+F23=‌
1
4πε0
‌
Q
a2
[‌
Q
2
+√2q
]

This force is directed along AC
(For the charge Q , at the point A , the force will have the same magnitude but will be directed along CA )
[Note : Don't deduct marks if the student does not write the direction of the net force, F]
(b) Potential energy of the system
‌=‌
1
4πε0
[4‌
qQ
a
+‌
q2
a√2
+‌
Q2
a√2
]

‌=‌
1
4πε0a
[4qQ+‌
q2
√2
+‌
Q2
√2
]

OR
(a) Finding the magnitude of the resultant force on charge q
(b) Finding the work done
(a) Force on charge q due to the charge −4q
F1=‌
1
4πε0
(‌
4q2
l2
)
‌, along
‌AB

Force on the charge q, due to the charge 2q
F2=‌
1
4πε0
(‌
2q2
l2
)
‌, along
‌CA

The forces F1 and F2 are inclined to each other at an angle of 120∘

Hence, resultant electric force on charge q
‌F=√F12+F22+2F1F2‌cos‌θ
‌=√F12+F22+2F1F2‌cos‌120∘
‌=√F12+F22−F1F2
‌=(‌
1
4πε0
‌
q2
l2
)
√16+4−8

‌=‌
1
4πε0
(‌
2√3q2
l2
)

(b) Net P.E. of the system
‌=‌
1
4πε0
â‹…
‌
q2
l
[−4+2−8]

‌=‌
(−10)
4πε0
‌
q2
l

‌∴‌ Work done ‌=‌
10q2
4pε0l
=‌
5q2
2πε0l
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