Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 2388 N/C. If the voltage is tripled and the distance between the plates is reduced to 1 5 the original distance, what is the magnitude of the new electric field

The magnitude of the new electric field is 35820 N/C.

Explanation:

Given:

Original magnitude of electric field (E₀) = 2388 N/C

Original voltage = 'V' (Assume)

Original separation between the plates = 'd' (Assume)

Now, new voltage is three times original voltage. So,

New distance is 1/5 the original distance. So,

Now, electric field between the parallel plates originally is given as:

Let us find the new electric field based on the above formula.

Now, . So,

Therefore, the magnitude of the new electric field is 35820 N/C.

v=ir = 3x15 =45 v

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2