The electric field between the plates of a parallel-plate capacitor is horizonta

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-2.25 HC 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 20° (a) What is the mass of the object? 14.57 Your response differs from the correct answer by more than 10%. Double check your calculations. g (b) What is the magnitude of the electric field? 7.189e-7X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N/C

Explanation / Answer

given

q = -2.25 micro C

T = 0.35 N

theta = 20 degrees

let m is the mass of the object and E is the magnitude of elctric field.

a) As the object is in equilibrium, net force acting on it must be zero.

Fnety = 0

T*cos(20) - m*g = 0

m = T*cos(20)/g

= 0.35*cos(20)/9.8

= 0.0336 kg

= 33.6 g

b) now use, Fnetx = 0

T*sin(20) - q*E = 0

E = T*sin(20)/q

= 0.35*sin(20)/(2.25*10^-6)

= 5.32*10^4 N/c

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