Four point charges, each of magnitude 5.04 µC, are placed at the corners of a sq

Question

Four point charges, each of magnitude
5.04 µC, are placed at the corners of a square
88.1 cm on a side.
If three of the charges are positive
and one is negative, find the magnitude
of the force experienced by the negative
charge. The value of Coulomb’s constant is
8.98755 × 109 N · m2/C2.
Answer in units of N

Explanation / Answer

All three positive charges will attract the negative charge Let the first particle be diagonally opposite => F1 = force on negative due to 1st particle = k* q*q /(v2 d)2 = 8.98755 × 109 * 5.04*5.04 *10-12/ ( 2 * 0.881*0.881) = 0.147 N this force will have a component in directions of both F2 & F3 and component will be F' = F1/v2= 0.104 N So along the direction of F2 , force on negative particle = k* q*q /(d)2 +F' = 8.98755 × 109 * 5.04*5.04 *10-12/ (0.881*0.881) + F' = 0.294 N + F' = 0.294 + 0.104 = 0.398 N Similarly along the direction of F3 , force on negative particle = k* q*q /(d)2 +F1/v2 = 0.398 N Final force will be vector sum of all these and in the direction of F1 F = [ (0.398)2 + (0.398)2 ] 1/2 = 0.563 N Hope this will do. :)

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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