Aparallel-plate capacitor is charged up. the magnitude of the charge on the square plates of the parallel-plate capacitor is q. the capacitor is then disconnected from the voltage supply, and the plates are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. the amount of charge on the plates is now equal to

The new value of the charge is just the same.

Explanation:

The capacitance of a parallel-plate capacitor, can be expressed in terms of its geometry and the properties of the dielectric material between the plates.

If the dielectric is air, we can approximate the formula for capacitance as follows:

C = ε₀ * A /d, where d is the distance between plates.

If d is made twice the original distance pulling apart the plates, the new value for C will be:

C₁= ε₀ * A / 2*d = C₀/2

By definition, C= Q/V.

As Q can´t change (due to it is disconnected from the battery), in order to do C₁ = C₀/2, the only choice must be V₁ = 2V₀, so Q₁=Q₀.

e/c, where e is its total energy

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft