Question

A positive source object of charge Qs and mass Ms is held fixed at the origin. A positive test object of charge qo and mo will feel an electrical force Fe when placed a particular distance, r, away. If we double the source charge to 2Qs then, what else can we do to make sure that test charge experiences the same electrical force, Fe, again?
A) Double the mass of the test object to 2 mo
B) Halve the charge of the test object to (1/2)q0 )
C) Increase the distance, r, between the test and source objects by a factor of 2
D) Increase the distance, r, between the test and source objects by a factor of √2
E) Either (B) or (D) will work.

Answer 1

Answer:

E) Either (B) or (D) will work

Step-by-step explanation:

In the first case with source positive charge Qs   and q₀

the electric force is:          K =  9 * 10⁹ Nm²C⁻²

Fe₁  =  K *  Qs*q₀ / r₁²      (1)

In the second case with source positive charge 2Qs  and  q₀

the electric force is:

Fe₂  =  K * 2*Qs*q₀ / r₂²

If the forces  Fe₁     and Fe₂   are to be equal then:

Fe₁   = Fe₂

K *  Qs*q₀ / r₁²  =   K * 2*Qs*q₀ / r₂²

1/r₁²  =  2/r₂²       or    r₂²   =  2* r₁²

r₂   =  √ 2 *  r₁

Then if we want to keep the same force we need to increase r₂ by a factor √2.

If we have  q₀₂   = 1/2 * q₀

Fe₃  =  K * 2*Qs* (1/2)*q₀  / r₁²

we can see that   Fe₃  =  K* Qs*q₀ / r₁² , and this expression is the same that   Fe₁  =  K *  Qs*q₀ / r₁² .

Again we can have the same electric force if we halve the charge q₀/2