# Electric force and its proportionality on charge(q)

Thread Starter

#### logearav

Joined Aug 19, 2011
243
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.

#### t_n_k

Joined Mar 6, 2009
5,455
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.

• logearav

#### timekeeper

Joined Oct 24, 2007
8
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:
• logearav
Thread Starter

#### logearav

Joined Aug 19, 2011
243
Thanks for your valuable replies t_n_k and timekeeper. It is so helpful to me

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