Question

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

Answer 1

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

Step-by-step 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,
`V_n=3V`

New distance is 1/5 the original distance. So,
`d_n=(d)/(5)`

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

`E_0=(V)/(d)=2388\ N/C`

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

`E_n=(V_n)/(d_n)\\\\E_n=(3V)/((d)/(5))\\\\E_n=15((V)/(d))`

Now,
`(V)/(d)=2388\ N/C`. So,

`E_n=15* 2388=35820\ N/C`

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