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

E'=(1/4)E

Step-by-step explanation:

The magnitude of the electric force is given by:

`E=k(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(q)/((2m)^2)=k(q)/(4m^2)`

when the distance is doubled we obtain:

`E'=k(q)/((4m)^2)=k(q)/(16m^2)=(1)/(4)k(q)/(4m^2)=(1)/(4)E`

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

hope this helps!

Answer 2

`E_n = (kQ)/(2) = 2E`

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

Step-by-step explanation:

Electric field due to the negative charge is given as:

`E = (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 = |(-kQ)/(2^2)| \\\\\\E = (kQ)/(4)`

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

`E_n = |(-2kQ)/(4)| \\\\\\E_n = |(-kQ)/(2)|`

`E_n = (kQ)/(2) = 2E`

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