39.Two positive charges 12micro Columbs and 10micro Columbs are initially separated by 10cm.The work done in bringing the two charges 4cm closer is
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Potential energy of the system = 1/(4πε) * Q1 Q2 / d
where ε = electrostatic permittivity
Q1 = charge = 12 μC
Q2 = 10 μC d = 10 cm = 0.1 meter
U1 = 9 * 10⁹ * 12 * 10⁻⁶ * 10 * 10⁻⁶ / 0.10
= 10.8 Joules
d2 = 6 cm
U2 = 9 * 10⁹ * 12 * 10⁻⁶ * 10 * 10⁻⁶ / 0.06
= 18 Joules
The work done by the external force agent in bringing together the charges from 10 cm to 6 cm is 18 - 10.8 J = 7.2 Joules
where ε = electrostatic permittivity
Q1 = charge = 12 μC
Q2 = 10 μC d = 10 cm = 0.1 meter
U1 = 9 * 10⁹ * 12 * 10⁻⁶ * 10 * 10⁻⁶ / 0.10
= 10.8 Joules
d2 = 6 cm
U2 = 9 * 10⁹ * 12 * 10⁻⁶ * 10 * 10⁻⁶ / 0.06
= 18 Joules
The work done by the external force agent in bringing together the charges from 10 cm to 6 cm is 18 - 10.8 J = 7.2 Joules
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