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

Answer:

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, $V_n=3V$

New distance is 1/5 the original distance. So, $d_n=\dfrac{d}{5}$

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

$E_0=\frac{V}{d}=2388\ N/C$

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

$E_n=\frac{V_n}{d_n}\\\\E_n=\frac{3V}{\frac{d}{5}}\\\\E_n=15(\frac{V}{d})$

Now, $\frac{V}{d}=2388\ N/C$. So,

$E_n=15\times 2388=35820\ N/C$

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