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
