Two points charges A and B of values +5 × 10^(-9) C and +3 × 10^(-9) C are kept 6 cm apart in air. When the charge B is moved by 1 cm towards charge A, then work done is equal to
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1
Answer:
as we know that +q and -q charges are lie on the z axis .so they form dipole and x and y axis are equatorial line so their potential is zero and the work done to moving the test charge from Q is also zero.
Answered by
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Initial distance between the charges Q
A
and Q
B
, d=5 cm =0.05 m
Potential energy of the system initially, U
i
=
d
kQ
A
Q
B
where k=9×10
9
∴ U
i
=
0.05
9×10
9
×3×10
−9
×1×10
−9
=5.4×10
−7
J
Now the charge B is moved by 1cm towards A.
Thus new distance between the charges d
′
=5−1=4 cm =0.04 m
Potential energy of the system initially, U
f
=
d
′
kQ
A
Q
B
∴ U
f
=
0.04
9×10
9
×3×10
−9
×1×10
−9
=6.75×10
−7
J
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