A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a
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Explanation:
It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.
So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.
U = qV
Since, the electric potential is directly proportional to the charge. Hence, when 0.2 tends to replaced by 0.4
then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.
Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.