Twice the amplitude of each wave
Answer:
the force will decrease to 3/4 of its original value.
Explanation:
The initial electric force between the two charges is:
![F = k \frac{q\cdot q}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20k%20%5Cfrac%7Bq%5Ccdot%20q%7D%7Br%5E2%7D)
where
k is the Coulomb's constant
q is the magnitude of each charge
r is their separation
Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of
![q+\frac{q}{2}=\frac{3}{2}q](https://tex.z-dn.net/?f=q%2B%5Cfrac%7Bq%7D%7B2%7D%3D%5Cfrac%7B3%7D%7B2%7Dq)
while the other charge will be
![q-\frac{q}{2}=\frac{q}{2}](https://tex.z-dn.net/?f=q-%5Cfrac%7Bq%7D%7B2%7D%3D%5Cfrac%7Bq%7D%7B2%7D)
So, the new force will be
![F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F](https://tex.z-dn.net/?f=F%27%20%3D%20k%20%5Cfrac%7B%28%5Cfrac%7Bq%7D%7B2%7D%29%5Ccdot%20%28%5Cfrac%7B3%7D%7B2%7Dq%29%7D%7Br%5E2%7D%3D%5Cfrac%7B3%7D%7B4%7D%20%28k%5Cfrac%7Bq%5Ccdot%20q%7D%7Br%5E2%7D%29%3D%5Cfrac%7B3%7D%7B4%7DF)
So, the force will decrease to 3/4 of its original value.
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