The electric field intensity at a point 2 m from an isolated point charge is 500 N/C. The electric potential at the point is
(A) 0 V
(B) 2.5 V
(C) 250 V
(D) 1000 V
Answered Feb 05, 2023
Correct option is: (D) 1000 V
Given: q = 10 µC = 10 × 10-6 C, F = 16 × 10-4 N To find: Electric field intensity (E) Formula: E = \(\frac{F}{q}\) Calculation: From formula, E = \(\frac{16\times10^{-4}}{10\times10^{-6}}\) = 160 N/C
VA = -10V, VB = 0, q = 2 × 10-6 C To Find: Work done (W) Formula: VBA = \(\frac{W}{q}\) Calculation: From formula, W = VBA × q = (VB – VA) × q = (0 + 10) × 2 × 10-6 = 20 ×...
(B) energy is used from some outside source
Correct option is: (B) [M1L1T-3A-1]
Correct option is: (A) 1500 NC-1
Correct option is: (A) 1.39 × 10-11 C
Correct option is: (D) 2 µC
(B) radially outward, radially inward
Correct option is: (A) 10 V
Correct option is: (B) pE sinθ
Install the Bissoy app to consult with a doctor.
Log in to ask questions, provide answers, or leave comments.