two charges -e and +e are X distance apart at waht distamce charge Q must be placed from charge +e that it is in equilibrium?
Answer (1)
21st Jun, 2020
Hello there,
It will be easier if you draw the diagram first. Read the question, draw the diagram accordingly. It will be helpful.
Now let's solve the problem:
Let us assume that,
Distance between Q and +e = d
Distance between Q and -e = X-d
Force of attraction due to -e = K*{(Qe)/X-d}
Force of attraction due to +e = K*{(Qe)/d}
Here, K= 1/(4Eo)
Now, these are in equilibrium. Then,
K*{(Qe)/X-d}= K*{(Qe)/d}
Or, X-d = d
Or, 2d = X
Or, d = X/2
Hence, the charge Q should be placed at X/2 position from +e, which means exactly in the middle of the two charges given initially.
Hope this answer helps you.
Thank you!!
It will be easier if you draw the diagram first. Read the question, draw the diagram accordingly. It will be helpful.
Now let's solve the problem:
Let us assume that,
Distance between Q and +e = d
Distance between Q and -e = X-d
Force of attraction due to -e = K*{(Qe)/X-d}
Force of attraction due to +e = K*{(Qe)/d}
Here, K= 1/(4Eo)
Now, these are in equilibrium. Then,
K*{(Qe)/X-d}= K*{(Qe)/d}
Or, X-d = d
Or, 2d = X
Or, d = X/2
Hence, the charge Q should be placed at X/2 position from +e, which means exactly in the middle of the two charges given initially.
Hope this answer helps you.
Thank you!!
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