Question

The magnitude of the electric field at a distance of two meters from a negative point charge is E. What is the magnitude of the electric field at the same location if the magnitude of the charge is doubled.

Answer 1

Answer:

E'=(1/4)E

Explanation:

The magnitude of the electric force is given by:

$E=k\frac{q}{r^2}$

where k is the Coulomb constant (8.89*10^{9}Nm^2/C^2).

When the distance is r=2m we have:

$E=k\frac{q}{(2m)^2}=k\frac{q}{4m^2}$

when the distance is doubled we obtain:

$E'=k\frac{q}{(4m)^2}=k\frac{q}{16m^2}=\frac{1}{4}k\frac{q}{4m^2}=\frac{1}{4}E$

Hence, the new electric field is a quarter of the first electric field.

hope this helps!

Answer 2

Answer:

$E_n = \frac{kQ}{2} = 2E$

If the charge is doubled, the electric field is also doubled.

Explanation:

Electric field due to the negative charge is given as:

$E = \frac{kQ}{r^2}$

where k = Coulomb's constant

Q = electric charge

r = distance between charge and point of consideration

At 2 m from the negative charge, the magnitude of the Electric field due to a negative charge -Q is given as E:

$E = |\frac{-kQ}{2^2}| \\\\\\E = \frac{kQ}{4}$

If the charge is doubled, the new charge becomes -2Q and the new electric field becomes:

$E_n = |\frac{-2kQ}{4}| \\\\\\E_n = |\frac{-kQ}{2}|$

$E_n = \frac{kQ}{2} = 2E$

If the charge is doubled, the electric field is also doubled.