Respuesta :
Answer:
3.974 Joule
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
[tex]U=\frac{kQq}{a}\\\Rightarrow U=\frac{9\times 10^9\times 5\times 10^{-6}\times 3.4\times 10^{-6}}{0.0385}=3.974\ J[/tex]
∴ Work to move a tiny 3.4 mC charge from very far away to the center of the ring is 3.974 Joule
Answer:
U = 3.97 J
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
[tex]WORK DONE U = \frac{kqQ}{a}[/tex]
=[tex]\frac{8.99*10^9*3.40*10^{-6} *5*10^{-6}}{3.85*10^{-2}}[/tex]
U = 3.97 J
