Question

Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. What is the electric field at \(x=d\) (point \(P\) )?

Short answer

Answer: The electric field at x=d is \(\frac{kq}{d^2}\).

Step-by-step solution

Calculate distances

Find the distance between point \(P\) and each charge. - Distance between the point charge \(q\) and \(P\): \(d_1=x_P-x_q = d-0=d\) - Distance between the point charge \(2q\) and \(P\): \(d_2=x_{2q}-x_P = 3d-d=2d\)

Calculate individual electric fields

Calculate the electric field generated by each charge at point \(P\) using the formula: \(E=\frac{kq}{r^2}\), where \(k\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance. - Electric field generated by charge \(q\) at point \(P\): \(E_1=\frac{kq}{d^2}\) - Electric field generated by charge \(2q\) at point \(P\): \(E_2=\frac{k(2q)}{(2d)^2}\)

Combine the electric fields

Since the charges are positive, the electric fields generated by both charges will push away from the charges. Therefore, electric field \(E_1\) will have a negative direction (towards the left), while electric field \(E_2\) will have a positive direction (towards the right). Thus, we should take their difference to find the net electric field at point \(P\). Total electric field at \(P\): \(E_P = E_2 - E_1\) Substitute the expressions for \(E_1\) and \(E_2\): \(E_P = \left(\frac{k(2q)}{(2d)^2}\right) - \left(\frac{kq}{d^2}\right)\) Simplify the expression: \(E_P = \frac{2kq}{4d^2}-\frac{kq}{d^2}\) \(E_P = \frac{kq}{d^2}\) So, the electric field at \(x=d\), or point \(P\), is \(\boxed{\frac{kq}{d^2}}\).