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Multiple Choice
A spherical balloon is charged so that the potential on its surface is . The balloon is now deflated to half its original radius (with no loss of charge). What is the new potential on its surface?
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Verified step by step guidance
1
Understand that the potential on the surface of a charged sphere is given by the formula: , where is Coulomb's constant, is the charge, and is the radius of the sphere.
Note that the charge remains constant as the balloon is deflated, and the initial potential is 40 V with the initial radius .
When the radius is halved, the new radius becomes . Substitute this new radius into the potential formula: .
Simplify the expression for the new potential: , which shows that the new potential is twice the original potential.
Since the original potential was 40 V, the new potential is V, which results in 80 V.