Question

A spherically symmetric charge distribution produces the electric field, E=(5000r²)rN/C, where r is in meters. What is the electric field strength at r=20 cm?

A. 4×10⁶ N/C
B. 2×10⁶ N/C
C. 5×10⁶ N/C
D. 1×10⁶ N/C

Answer 1

The electric field strength at a distance of 20 cm from a spherically symmetric charge distribution that generates an electric field E=(5000r²)rN/C is found to be 40 N/C after substituting 0.20 meters for r.

Step-by-step explanation:

To determine the electric field strength at a given distance r from a spherically symmetric charge distribution, which produces the field E=(5000r²)rN/C, we'll first convert the distance to meters, as the equation uses SI units. For r=20 cm, which is 0.20 meters, the calculation is straightforward.

Substitute the value of r into the given electric field equation:

E = 5000 × (0.20)² × (0.20)
E = 5000 × 0.04 × 0.20

E = 5000 × 0.008E = 40 N/C

Thus, the electric field strength at r=20 cm is 40 N/C.