Question

A charge of +1 Coulomb is place at the 0- cm mark of a meter stick. A charge of −1 Coulomb is placed at the 100-cm mark of the same meter stick. Is it possible to place a proton somewhere on the meter stick so that the net force on it due to the two charges is 0? 1. Yes; to the left of the 50-cm mark 2. Yes; to the right of the 50-cm mark 3. No

Answer 1

It is not possible. See explanation below

Step-by-step explanation:

The distribution of charges makes that the net force acting on the proton is always different from zero if we want to place it in between the charges. This is because the forces act in the same direction and therefore the vectors always add. See the attached sketch for the case when the proton is to the left of the 50 cm mark. The same can be applied for the case in which it is placed to the right.

It is important to note that we could have also try to place the proton either to the right of the negative charge or to the left of the positive charge. In these cases, the forces have opposite directions and it could be possible to obtain a net force equals to zero. Using the superposition principle for the Coulomb force we have:

`\displaystyle{F_(net)=(qp)/(4\pi\epsilon_0r_+^2)-(qp)/(4\pi\epsilon_0r_-^2)}`

but we want this force to be zero, thus:

`\displaystyle{(qp)/(4\pi\epsilon_0r_+^2)=(qp)/(4\pi\epsilon_0r_-^2)}}`

and since the charges are the same:

`\displaystyle{(1)/(r_+^2)=(1)/(r_-^2)}}`

which is a contradiction because the distances are different.