Example – 1 Three charges, –q, Q, q are placed at equal distances on a straight line. If the total potential energy of the system of the three charges is zero, then Q : q = .....
Solution : Let d be the equal distance. The total potential energy of the system is, U = U12 + U23 + U31
Since U = 0
-2Q + (q/2) = 0 or , -4Q + q = 0 or, 4Q = q or, (Q/q) = 1 : 4
Example – 2 Three charges are arranged as shown in fig. Find the potential energy of the system.
Solution : The potential energy of the system is U = U12 + U23 + U31
or U = 9 × 109 × 10-13 (-10)
= -9 × 109 × 10-132 Þ -9 × 10-13joules
Example – 3 An electron (charge, -e) is placed at each of the eight corners of a cube of side a and a- particle (charge, +2e) at the centre of the cube. Compute the P.E of the system.
Solution : Fig., shows an electron placed at each to the eight corners of the cube and a particle at the centre. The total energy of the system is the sum of energies of each pair of charges. There are 12 pairs like A and B (separation a), 12 paris like A and C (separation,
Ö2a ), 4 pairs like A and G (separation Ö3a ) and 8 paris like A and 0
(separation (Ö3/2)a).
Hence
= (9 x 109) (e2/a) (4.32) = 3.9 x 1010 (e2/a) J.
Example – 4 The value of q1 and q2 are 2 × 10-8 coulomb and 0.4 × 10-8 coulomb respectively as shown in fig. A third charge q3 = 0.2 × 10-8 coulomb is moved from point C to point D along the arc of a circle. The change in the potential energy of charge will be –
Solution : Potential energy of q3 at point C
Potential energy of q at point D
\ UD - Uc = kq2q3 [(1/0.2)-(1/1)]
= 9 × 109 × 0.4 × 10-8 × 0.2 × 10-8 × 4
= 2.88 × 10-7 joule
Note : Note Charge potential energy charge will ne same even if we had taken
any other path as electric field is conservative.)
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