Two positive point charges Q = +4.00 C and q = +2.00C are placed at the opposite

Question

Two positive point charges Q = +4.00 C and q = +2.00C are placed at the opposite corners of a rectangle
with a length of 0.800m and a width of 0.400m. Charge Q is placed at the top right corner of the rectangle and charge q is placed at the left bottom corner of the rectangle, Point A is at the top left empty corner and point B is at the right bottom empty corner. What are the potentials at points A and B (relative to infinity) due to these charges?

(1) 8.90 × 104 V at A and 8.90 × 104 V at B
(2) 1.12 × 105 V at A and 8.90 × 104 V at B
(3) 8.90 × 104 V at A and 1.12 × 105 V at B
(4) 0.00V at A and 1.12 × 105 V at B
(5) None of the above.

Explanation / Answer

Electric potential is a scalar quantity, which is given by:

V = kQ/R

Va = V1 + V2

Va = kQ/r1 + k*q/r2

r1 = 0.8 m

r2 = 0.4 m

Va = 8.99*10^9*(4*10^-6/0.8 + 2*10^-6/0.4)

Va = 8.9*10^4 V

Potential at B will be

Vb = V1 + V2

Vb = kQ/r1 + k*q/r2

r1 = 0.4 m

r2 = 0.8 m

Vb = 8.99*10^9*(4*10^-6/0.4 + 2*10^-6/0.8)

Vb = 1.12*10^5 V

Correct option is 3.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.