Electric force and its proportionality on charge(q)

Thread Starter


Joined Aug 19, 2011
The Electric field due to a source charge Q is defined in terms of some test charge q. Electric force F is proportional to q. Upto this i understood. Now the next line of my book says, Since F is proportional to q, the ratio F/q does not depend on q..
I dont understand the text highlighted. Why F/q does not depend on q?
Revered members, please throw light on my query.


Joined Mar 6, 2009
Well if F is directly proportional to 'q' then F=K*q, where K is some constant of proportionality.

Hence the ratio F/q=(K*q)/q=K, which being a constant is independent of q.


Joined Oct 24, 2007
Let Q be the charge which causes an electric field, q1 the test charge and r the distance between them. The Force that q1 experiences at distance r because of Q is:
\(F(q1)=K\frac{Q*q1}{r^2}\) This is your first equation.

Now take a different test charge q2 and put it at the same distance r again. For q2 the Force because of Q is:
\(F(q2)=K\frac{Q*q2}{r^2}\) This is your second equation.

Now, notice that \(\frac{F(q1)}{q1}\) is equal to \(\frac{F(q2)}{q2}\) !!! And since we define \(E=\frac{F(q)}{q} = \frac{K\frac{Q*q}{r^2}}{q} \Rightarrow E= K\frac{Q}{r^2}\)
As you can see, we have reached to an equation which considers only source charge Q and distance r from Q.

The Electric field of a given charge Q at a distance r from it is always the same, no matter what test charge q you put there.
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