A complete charge Q = 3.6 μC is spread uniformly over a 4 minutes 1 circle arc the radius a = 8.8 cm together shown.

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a) What is λ the direct charge density along the arc?

b) What is Ex, the value of the x-component the the electric field in ~ the beginning (x,y) = (0,0) ?

c) What is Ey, the value of the y-component that the electric field at the beginning (x,y) = (0,0) ?


Answer

a) The direct charge density is expressed as follows:

lambda=fracQL

Convert the charge from micro coulombs to coulombs.

eginalignedQ=(3.6 mu mathrmC)left(frac10^-6 mathrmC1 mu mathrmC ight) \=3.6 imes 10^-6 mathrmCendalignedConvert the radius indigenous centimeter come meters.

eginaligneda=(8.8 mathrm~cm)left(frac10^-2 mathrm~m1 mathrm~cm ight) \=8.8 imes 10^-2 mathrm~mendalignedSubstitute fracpi2 a for L in expression lambda=fracQL.


lambda=fracQfracQ2 a

Substitute 8.8 imes 10^-2 mathrm~m for a and 3.6 imes 10^-6 mathrmC for a in expression lambda=fracQfracpi2 a.

eginalignedlambda=frac3.6 imes 10^-6 mathrmCfracT2left(8.8 imes 10^-2 mathrm~m ight) \=2.60 imes 10^-5 mathrmC / mathrmmendalignedPart a) The linear charge thickness is 2.6 imes 10^-5 mathrmC / mathrmm.

b) The electrical field at beginning in mathrmx direction is expressed as follows:

E_x=-int_0^fracpi2 fraclambda cos heta4 pi varepsilon_0 a d heta

Solve the integration to calculation the expression to calculation the electric field.

eginalignedE_x=-fraclambda4 pi varepsilon_0 a_0^fracpi2 \=-fraclambda4 pi varepsilon_0 aleft(sin left(fracpi2 ight)-sin (0) ight) \=-fraclambda4 pi varepsilon_0 a(1-0) \=-fraclambda4 pi varepsilon_0 aendalignedSubstitute 2.6 imes 10^-5 mathrmC / mathrmm for lambda, 9.0 imes 10^9 fracmathrmN cdot mathrmm^2mathrmC^2 because that frac14 pi varepsilon_0, and also 8.8 imes 10^-2 mathrm~m because that a in expression E_x=-fraclambda4 pi varepsilon_0 a.

E_x=-left(9.0 imes 10^9 fracmathrmN cdot mathrmm^2mathrmC^2 ight)left(frac2.6 imes 10^-5 mathrmC / mathrmm8.8 imes 10^-2 mathrm~m ight)

=-2.6 imes 10^6 mathrm~N / mathrmC

The x - ingredient of the electrical field is command along an adverse x - direction.

Part b) The worth of the x-component the the electrical field at the beginning is -2.6 imes 10^6 mathrm~N / mathrmC.

(c) The electrical field at origin in y direction is expressed as follows:

E_y=-int_0^fracpi2 fraclambda sin heta4 pi varepsilon_0 a d heta

Here, lambda is the direct charge density, a is the radius of arc, heta is the angle made by the arc, varepsilon_0 is the permittivity of free space, and also E_y is the electric field in y direction. Combine the expression to find the expression of electric field in upright direction.

eginalignedE_y=-left(-int_0^fracpi2 fraclambda4 pi varepsilon_0 a_0^fracpi2 ight) \=fraclambda4 pi varepsilon_0 aleft(cos left(fracpi2 ight)-cos (0) ight) \=-fraclambda4 pi varepsilon_0 a(1) \=-fraclambda4 pi varepsilon_0 aendalignedSubstitute 2.6 imes 10^-5 mathrmC / mathrmm for lambda, 9.0 imes 10^9 fracmathrmN cdot mathrmm^2mathrmC^2 because that frac14 pi varepsilon_0, and 8.8 imes 10^-2 mathrm~m because that a in expression E_y=-fraclambda4 pi varepsilon_0 a.

eginalignedE_y=-left(9.0 imes 10^9 fracmathrmN cdot mathrmm^2mathrmC^2 ight)left(frac2.6 imes 10^-5 mathrmC / mathrmm8.8 imes 10^-2 mathrm~m ight) \=-2.6 imes 10^6 mathrm~N / mathrmCendalignedThe y - ingredient of the electric field is command along an adverse y-direction.

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Part c) The worth of the y-component the the electrical field at the beginning is -2.6 imes 10^6 mathrm~N / mathrmC.