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, [tex]V_n=3V[/tex]
New distance is 1/5 the original distance. So, [tex]d_n=\dfrac{d}{5}[/tex]
Now, electric field between the parallel plates originally is given as:
[tex]E_0=\frac{V}{d}=2388\ N/C[/tex]
Let us find the new electric field based on the above formula.
[tex]E_n=\frac{V_n}{d_n}\\\\E_n=\frac{3V}{\frac{d}{5}}\\\\E_n=15(\frac{V}{d})[/tex]
Now, [tex]\frac{V}{d}=2388\ N/C[/tex]. So,
[tex]E_n=15\times 2388=35820\ N/C[/tex]
Therefore, the magnitude of the new electric field is 35820 N/C.
