if some charges are placed on x axis at x=0, x=1, x=2,x=4,x=8(in cm)and so on. . find net force on charge at 1 cm if each charge is of 2 micro coulomb.
is it 0 n (like charge repels)

Is answer to the above question is 0 N.(CONCEPT- LIKE CHARGES REPELS)

Please clarify - are the values along the x-axis in centimeters? If so, then the charge at x=1 cm it will be repelled by the same magnitudes of force from charges at x = 0 and x=2, as each of these are 1 cm away from x=1 and opposing forces, which cancel. That leaves a net force from the charges at x = 4, x = 8, x=16, etc, which are distances of 3, 7, 15, etc cm away. The magnitude of those forces is proportional to the inverse of the distance squared. So determining net force acting on a charge at x=1 will involve calculating the infinite sum of (1/3)^2 + (1/7)^2 + (1/15)^2 + ....

Please clarify - are the values along the x-axis in centimeters? If so, then the charge at x=1 cm it will be repelled by the same magnitudes of force from charges at x = 0 and x=2, as each of these are 1 cm away from x=1 and opposing forces, which cancel. That leaves a net force from the charges at x = 4, x = 8, x=16, etc, which are distances of 3, 7, 15, etc cm away. The magnitude of those forces is proportional to the inverse of the distance squared. So determining net force acting on a charge at x=1 will involve calculating the infinite sum of (1/3)^2 + (1/7)^2 + (1/15)^2 +...

SIR,
THANK YOU.
BUT STILL I HAVE A DOUBT, WON'T THE FORCES EXERTED BY CHARGES OF 2 MICRO COULOMBS AT X= 4 cm,8 cm... ON THE CHARGE AT X=1 cm CANCELS OUT AND BECOME 0 N DUE TO THE PRESENCE OF CHARGE AT X=0 AS THE CHARGES ARE ELECTROSTATIC IN NATURE OR IS IT (2*2)/3^2+(2*2)/7^2 AND SO ON( coulombs law)

Apply Coulomb's Law, which says the force between any two charges is proportional to the inverse distance squared, and the total force acting on any one charge is the sum of all forces acting on it. So the forces acting on the charge at x=1 from the charge at x=0 and x=2 cancel out, but there is nothing to cancel the forces acting on x=1 from the charges at x= 4, x=8, x=16 etc.

SIR ,
Thank you. But still i have a doubt.(the force on x=4cm ,8cm on x=1cm do not cancel out) is it because of the varying distance.
Then will the net force acting on x=1cm be ( 2*2)/3*3+(2*2)/7*7 and so on based on coulombs law.

sir ,
thank you. But still i have a doubt.(the force on x=4cm ,8cm on x=1cm do not cancel out) is it because of the varying distance.
Then will the net force acting on x=1cm be ( 2*2)/3*3+(2*2)/7*7 and so on based on coulombs law.is it true or not.

Then will the net force acting on x=1cm be ( 2*2)/3*3+(2*2)/7*7 and so on based on coulombs law.

Yes, as I explained back in post #3 of this thread.

yes, as i explained back in post #3 of this thread.


sir,
still one more doubt . Isn't ( 1/3)^2+(1/7)^2... Different from (2*2)/3*3+(2*2)/7*7.
^ means exponent ,but why (1/3)^2 instead of (2/3)^2 ,as charges on every body is 2 micro coulomb,right.

I was describing the effect of distance only, to show that the forces don't cancel out. However, since you asked, the actual equation would be:

F = k_e(2 \times 10^{-6} C)^2 [(\frac 1 {0.03 m})^2 + (\frac 1 {0.07m})^2 + (\frac 1 {0.15m})^2 + ...]

where k_e is Coulomb's constant in units appropriate for charge expressed in coulombs and distance measured in meters.

I think this infinite sum is going to be difficult to evaluate. This problem would be a lot easier if it asked for the the sum of forces acting on the charge at x=0, as the resulting infinite sum is pretty straight-forward to evaluate.