Positive charge Q is uniformly distributed around a semicircle of radius a (see

Question

Positive charge Q is uniformly distributed around a semicircle of radius a (see figure). Find the electric field (magnitude and direction) at the center of curvature P. (Use the following as necessary: Q, k, and a.) upward to the right to the left downward 6" 1 points YF1422·P.033. My Notes Ask Your Teach + A small sphere with a mass of 0.004 g and carrying a charge of 5.20 x 10-8 C hangs from a thread near a very large, charged conducting sheet, as shown in the figure below The charge density on the sheet is 1.50 x 10-9 C/m2. Find the angle of the thread

Explanation / Answer

consider a small length dr having charge dq at an angle theta

dr = a*dtheta

dq = (Q/(pi*a))*dr = Q*dtheta/pi

electric field due to dq dE = -k*dq/a^2

along x axis

dEx = dE*sintheta

Ex = integral dEx from theta -90   to theta = 90

Ex = 0

dEy = -dE*costheta

dEy = -k*Q*costheta*dthete/(pi*a^2)

Ey = integration dEy from theta = -pi/2 to theta = pi/2

Ey = -(k*Q/(pi*a^2)) (sintheta) from theta = -pi/2 to theta = pi/2

Ey = -(k*Q/(pi*a^2)) *(1 - (-1))

Ey = -2*k*Q/(pi*a^2)

magnitude Ey = 2kQ/(pi*a^2)

direction = downward

=============================

electric field due to sheet E = sigma/(2*e0)

electric force on charge Fx = -E*q

In equilibrium

along horizontal

Fnet = 0

Fx + Tx = 0

-E*q + T*sintheta = 0

T*sintheta = Eq

along vertical

Fnet = 0

T*costheta - mg = 0

T*costheta = mg

tantheta = Eq/mg

tantheta = sigma*q/(2*e0*m*g)

tantheta = 1.5*10^-9*5.2*10^-8/(2*8.85*10^-12*0.004*10^-3*9.8)

theta = 6.41

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