Question

The electric field between the plates of a parallel-plate capacitor is horizontal, uniform and has a magnitude E. A small object with a charge of 2.10 μC is attached to the string. Assume that the tension in the string is 0.350 N, and the angle it makes with the vertical is 15 (b) What is the magnitude of the electric field? N/C search

Answer 1

Assuming the system is in equilibrium.

Then the Net force on the particle is zero.

Equating both X and Y component of the force to zero we get following

Assuming the system is in equilibrium. Then the Net force on the particle is zero. Equating both X and Y component of the force to zero we get following T cos (theta) = mg T sin (theta) = q E Thus, mass of the object is given by m = T cos (theta)/g = 0.35 times cos (15)/9.8 m = 0.0345 kg = 34.5 g Electric field is given by E = T sin (theta)/q = 0.35N times sin(15)/2.1 times 10^-6C E = 0.0431 times 10^6 N/C = 4.31 times 10^4 N/C

Thus, mass of the object is given by

$m = \frac{Tcos(\theta)}{g} = \frac{0.35*cos(15)}{9.8} \\ \\ m = 0.0345 kg = 34.5 g$

Electric field is given by

$E = \frac{Tsin(\theta)}{q} = \frac{0.35N*sin(15)}{2.1*10^{-6}C} \\ \\ E = 0.0431*10^6 N/C= 4.31*10^4 N/C$