Question

a point charge q of mass 8.50 g hangs from the horizontal ceiling by a light 25.0-cm thread. when a horizontal electric field of magnitude 1750 n/c is turned on, the charge hangs away from the vertical as shown in the figure. the magnitude of q is closest to

Answer 1

Answer:

27.5μC

Explanation:

Given:

m=8.5*10^-3

E=1750N/C

Fg=9.8m/s^2

θ to vertical = 30°

Find:

Q

Fg=mg=8.5*10^-3 * 9.8 = 0.0833N

Tcos(30)=mg -> T=(mg)/cos(30) = 0.096N

Fe(force of electric field on charge) = Tsin(30) = 0.096sin(30) = 0.048N

E=F/Q -> Q=Fe/E = 0.048/1750 = 27.5μC

(see attachment)

Answer 2

We have that the the magnitude of charge Q is mathematically given as

Q=27.5uC

Magnitude of charge

Question Parameters:

a point charge q of mass 8.50 g hangs from the horizontal ceiling by a light 25.0-cm thread.

when a horizontal electric field of magnitude 1750 n/c is turned on

Generally the equation for the Tension  is mathematically given as

For Horizontal Direction

$Tsin \tTsin \theta=Q\theta=QE$

For Vertical Direction

$Tcos\theta=mg$

$tan\theta=\frac{QE}{mg}$

Hence

$Sin \theta=\frac{r}{e}\\\\\theta=sin^{-1}(12.5/25)=30\\\\And\\\\tan \theta=\frac{QE}{mg}$

$Q=\frac{mg+tan\theta}{E}\\\\Q=\frac{8.5e-3*9.8*tan \theta 30}{1750}$

Q=27.5uC

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