answer questions c-f for this problem (25%) Problem 2: A positive charge of magn

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answer questions c-f for this problem

(25%) Problem 2: A positive charge of magnitude Q,-3.5 nC is located at the onigin. A negative charge 0--8.5 nC is located on the positive x-axis at x-4.5 cm from the origin. The point P is located y 4.5 cm above charge Q2 Qi 02 Ctheexpertta.com i 17% Part (a) Calculate the x component of the electric field at point p due to charge 2: Write your ans wer in units ofNC e' 17% Part (b) Calculate the y component of the electric field at point p due to charge write your answer in units of Ne 17% Part (c) Calculate the y component of the electric field at point p due to the Charge Q2 write your answer in units onc Grade Summary Deductions i5% Potential 82% Submissions Attempts remining cotan)asin acos0 atan) acotan sinho cosho tanh cotanh detailed vin DegreesRadians

Explanation / Answer

coordinates are:

Q1: (0,0)

Q2: (4.5*0.01,0) m

P:(4.5*0.01,4.5*0.01) m

field at P due to Q1:

as Q1 is positive, field will be away from Q1 and towards P.

field direction vector=(4.5*0.01,4.5*0.01)-(0,0)=(0.045,0.045)

distance=sqrt(2)*0.045=0.06364 m

unit vector along the field=(0.045,0.045)/0.06364=(0.707,0.707)

field magnitude=k*Q1/distance^2

where k=coloumb’s constant=9*10^9

field magnitude=9*10^9*3.5*10^(-9)/0.06364^2

=7777.7 N/C

in vector notation, field due to Q1=E1=7777.7*(0.707,0.707) N/C

field at P due to Q2:

as Q2 is negative, field will be away from P and towards Q2.

field direction vector=(4.5*0.01,0)-(0.045,0.045)=(0,-0.045)

distance=0.045 m

unit vector along the field=(0,-0.045)/0.045=(0,-1)

field magnitude=k*Q2/distance^2

where k=coloumb’s constant=9*10^9

field magnitude=9*10^9*8.5*10^(-9)/0.045^2

=3.7778*10^4 N/C

in vector notation, field due to Q2=E2=3.7778*10^4*(0,-1) N/C

answers are:

part c:

y component of field at P due to Q2=3.7778*10^(4)*(-1)=-3.7778*10^4 N/C

part d:

total field=E1+E2=(5.4988*10^3,-3.228*10^4) N/C

so y component=-3.228*10^4 N/C

part e:

magnitude of total field=sqrt((5.4988*10^3)^2+(3.228*10^4)^2)=3.2745*10^4 N/C

part f:

angle in degree=arctan(y component / x component)

=arctan((-3.228*10^4)/(5.4988*10^3))

=-80.333 degrees

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