# Electric force and its proportionality on charge(q)

Discussion in 'Physics' started by logearav, Jan 5, 2012.

1. ### logearav Thread Starter Member

Aug 19, 2011
248
0
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.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
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.

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3. ### timekeeper New Member

Oct 24, 2007
4
1
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.

Last edited: Jan 6, 2012
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4. ### logearav Thread Starter Member

Aug 19, 2011
248
0
Thanks for your valuable replies t_n_k and timekeeper. It is so helpful to me