Question

The electric potential ( relative to infinity ) due to a single point charge Q is 400 V at a point that is 0.6 m to the right of Q. The electric potential (relative to infinity) at a point that is 0.90 m to the left of 0 is:_____.

A. + 400 V.
B. -400 V.
C. + 200 V.

Answer 1

The electric potential at a point due to a single point charge is given by V = kQ/r. Using this formula, we can calculate the electric potential at a point 0.90 m to the left of the charge to be 2.4 x 10^12 V/m.

Step-by-step explanation:

The electric potential at a point due to a single point charge is given by:

V = kQ/r

Where V is the electric potential, k is the electrostatic constant, Q is the charge, and r is the distance from the charge.

In this problem, the electric potential at a point 0.6 m to the right of the point charge is 400 V. To find the electric potential at a point 0.90 m to the left of the charge, we can set up the following equation:

400 V = kQ/(0.6 m)

Solving for Q, we find that Q = 240 C.

Now, using the found value of Q and the distance 0.90 m, we can calculate the electric potential at this point:

V = kQ/(0.90 m)

Plugging in the values, we get:

V = (9.0 x 10^9 N m^2/C^2)(240 C)/(0.90 m) = 2.4 x 10^12 V/m

So, the electric potential at a point that is 0.90 m to the left of the point charge is 2.4 x 10^12 V/m.

Answer 2

The potential at a distance of 0.9 m is 266.67 V.

Step-by-step explanation:

Charge = Q

Potential is 400 V at a distance 0.6 m .

Let the potential is V at a distance 0.9 m.

Use the formula of potential.

`V = (Kq)/(r)\\\\(V)/(400)=(0.6)/(0.9)\\\\V = 266.67 V`