A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the center of the ring. It is observed that the electric field at point P, which is on the axis of the ring and 70 cm from its center, is equal to 2000 N/C and is directed away from the center of the ring. Determine the value of q.

Question

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Respuesta :

rejkjavik rejkjavik · Answer 1 · 2019-10-03T12:38:24+02:00

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

[tex]E = \frac{KQx}{[x^2+ R^2]^{3/2}}[/tex]

[tex]E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}[/tex]

E1 = 3714.672 N/C

electric field due to point charge q

[tex] E =\frac[kq}{x^2}[/tex]

[tex]E = \frac{9\times 10^9 \times q}{0.70^2}[/tex]

[tex]E2 = 1.837\times 10^{10}\times q[/tex]

now the eelctric charge at point P is

E = E1 + E2[tex]2000 = 3714.672 + 1.837\times 10[10} \times q[/tex]

solving for q

q = - 93.334 nC