I think i used calulater and it gives me 47.5
Answer:
6.44 × 10^10 N/C
Explanation:
Electric field due to the ring on its axis is given by
E = K q r / (r^2 + x^2)^3/2
Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.
r = 0.25 m, x = 0.5 m, q = 5 C
K = 9 × 10^9 Nm^2/C^2
E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2
E = 6.44 × 10^10 N/C
Answer: Not 100% sure but I think it’s C.
Hope this helps! ^^
Answer:
Yes
Explanation:
Given that the battery is the same the PD ( potential difference ) in the circuit will also be the same likewise the flow of charge in the circuit,
Hence the same amount of charge flow is delivered to any circuit.
attached below are examples