Four positive charges are fixed at the corners of a square, as shown above. Three of the charges have magnitude Q, and the fourth charge has a magnitude 2Q. Point P is at the center of the square at a distance r from each charge.

Question

Four positive charges are fixed at the corners of a square, as shown above. Three of the charges have magnitude Q, and the fourth charge has a magnitude 2Q. Point P is at the center of the square at a distance r from each charge.

22. What is the electric potential at point P ? (A) Zero
(B) kQ r
(C) 2kQ r
(D) 4kQ r
(E) 5kQ

Answer 1

It is so simple: I suspect you are over thinking.

Potential is a scalar, not vector. So add the four potentials, three are the same, and one is double, (2+3)

V=5k(q/r)

Answer 2

So the answer is E? or V=5k(q/r)

Answer 3

To find the electric potential at point P, we can use the concept of superposition. The electric potential at a point is the sum of the potentials due to all the charges.

First, let's calculate the potential at point P due to each charge individually.

1) Charge Q at the top left corner:
The distance between this charge and point P is r (as given). The electric potential due to a point charge is given by V = k * Q / r, where k is the electrostatic constant. Therefore, the potential at point P due to this charge is kQ / r.

2) Charge Q at the top right corner:
The distance between this charge and point P is also r. So, the potential at point P due to this charge is also kQ / r.

3) Charge Q at the bottom right corner:
The distance between this charge and point P is diagonal to the square, which can be calculated using the Pythagorean theorem. The diagonal of a square is √2 times the length of one side. Therefore, the distance is r√2. Hence, the potential at point P due to this charge is kQ / (r√2).

4) Charge 2Q at the bottom left corner:
The distance between this charge and point P is also diagonal to the square, which is again r√2. Using the formula, the potential at point P due to this charge is 2kQ / (r√2).

Now, we can add up the potentials due to all the charges:

V(total) = (kQ / r) + (kQ / r) + (kQ / (r√2)) + (2kQ / (r√2))

Let's simplify this expression:

V(total) = (2kQ / r) + (3kQ / (r√2))

We can combine the terms:

V(total) = (2√2kQ + 3kQ) / (r√2)

Simplifying further:

V(total) = (2√2 + 3)kQ / (r√2)

So, the electric potential at point P is ((2√2 + 3)kQ) / (r√2).

Comparing this with the given answer choices, none of the options match exactly. Therefore, the correct answer is not among the options provided.

Four positive charges are fixed at the corners of a square, as shown above. Three of the charges have magnitude Q, and the fourth charge has a magnitude 2Q. Point P is at the center of the square at a distance r from each charge.

It is so simple: I suspect you are over thinking.

So the answer is E? or V=5k(q/r)

To find the electric potential at point P, we can use the concept of superposition. The electric potential at a point is the sum of the potentials due to all the charges.

(b)babi