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A ring of diameter 7.70 cm is fixed in place and carries a charge of 5.00 mC uniformly spread over its circumference. (a) How much work does it take to move a tiny 3.40@mC charged ball of mass 2.50 g from very far away to the center of the ring

User Ramtin Gh
by
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2 Answers

2 votes

Answer:

U = 3.97 J

Step-by-step explanation:

GIVEN DATA:

a = 3.85 cm

q = 3.40 mc

q =5 mc

Work done to move charge to centre of the ring is equal to potential energy of the system


WORK DONE U = (kqQ)/(a)

=
(8.99*10^9*3.40*10^(-6) *5*10^(-6))/(3.85*10^(-2))

U = 3.97 J

User Michaelok
by
5.9k points
5 votes

Answer:

3.974 Joule

Step-by-step explanation:

Diameter of ring = 7.7 cm

a = Distance from the center = d/2 = 3.85 cm = 0.0385 m

Q = Charge = 5 mC

q = Charge to move = 3.4 mC

k = Coulomb constant = 9×10⁹ Nm²/C²

Work done will be equal to Potential energy when mass is at center


U=(kQq)/(a)\\\Rightarrow U=(9* 10^9* 5* 10^(-6)* 3.4* 10^(-6))/(0.0385)=3.974\ J

∴ Work to move a tiny 3.4 mC charge from very far away to the center of the ring is 3.974 Joule

User FBronner
by
4.8k points