Answer:
a. E(r) = 3.67 x 10⁴ N/C
b. E(r) = 5.38 x 10⁴ N/C
Explanation:
r1 = 8.83 cm
r2 = 15.4 cm
Charge on the inner shell = 5.03 x 10^-⁸ C
Charge on the outer shell = 2.77 x 10^-⁸ C
a) To find the electric field at r = 11.1 cm,
r₁ = 8.83 cm < r = 11.1 cm < r₂ = 15.4 cm
E(r) = <u>1 </u> <u>q₁ </u>
4∈₀ r₂
E(r) = <u>( 8.99 x 10⁹ N.m²/C² ) ( 5.03 x 10⁻⁸)</u>
(0.111m)²
E(r) = 3.67 x 10⁴ N/C
b) To find the electric field at r = 36.1 cm
since r₁ < r₂ < r = 36.1 cm
E(r) = <u>1 </u> <u>q₁ ₊ q₂ </u>
4∈₀ r₂
E(r) = <u>( 8.99 x 10⁹ N.m²/C² ) ( 5.03 + 2.77 )(1 x 10⁻⁸)</u>
(0.361m)²
E(r) = 5.38 x 10⁴ N/C