An infinite line of charge with charge density ? 1 = 4.8 ?C/cm is aligned with t

Question

An infinite line of charge with charge density ?1 = 4.8 ?C/cm is aligned with the y-axis as shown.

1) What is Ex(P), the value of the x-component of the electric field produced by by the line of charge at point P which is located at (x,y) = (a,0), where a = 6.9 cm? ANS: 125217000 N/C

2) What is Ey(P), the value of the y-component of the electric field produced by by the line of charge at point P which is located at (x,y) = (a,0), where a = 6.9 cm? ANS: 0

3) A cylinder of radius a = 6.9 cm and height h = 11.4 cm is aligned with its axis along the y-axis as shown. What is the total flux ? that passes through the cylindrical surface? Enter a positive number if the net flux leaves the cylinder and a negative number if the net flux enters the cylnder.

4) Another infinite line of charge with charge density ?2 = -14.4 ?C/cm parallel to the y-axis is now added at x = 3.45 cm as shown. What is the new value for Ex(P), the x-component of the electric field at point P?

5) What is the total flux ? that now passes through the cylindrical surface? Enter a positive number if the net flux leaves the cylinder and a negative number if the net flux enters the cylnder.

6) The initial infinite line of charge is now moved so that it is parallel to the y-axis at x = -3.45cm. What is the new value for Ex(P), the x-component of the electric field at point P?

7) The initial infinite line of charge is now moved so that it is parallel to the y-axis at x = -3.45cm. What is the new value for Ex(P), the x-component of the electric field at point P?

Explanation / Answer

3) Flux = E. Area perpendicular to it

=125217000 N/C* pi a2h = 125217000 *2*3.14*0.069  * 0.114 =6185529 Nm2/C

4) E is directly proportionalto charge density annd inversally proportional to distance,

here charge density becomes -3 times and distance becomes (3.45/6.9=1/2) half

So E due to negative line charge is -3 *2 = -6 times

So total E = Enegative + E positive =-6*E positive +E positive=-5*E positive

=-5*125217000 = -626085000 N/C

5) Using gauss law, total flux = Charge enclosed/ e

total charge enclosed become -2 times as (4.8-14.4=-9.6)

so flux becomes - 2 times of tht calculated in part 3

flux =-12371059 Nm2/C

6) Since initial charge goes away 3.45cm further, its distance becomes1.5 times, so Eduee to this becomes 1/1.5=0.667 times

E=-6*E positive +0.667E positive=-5.333*E positive =-667819826 N/C

7) it is same question as 6

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