Electric Fields: Help Needed for Questions on Charge and Magnitude

[swami85]

help needed on a couple questions - electric fields

anyone able to lend a hand?

1. How many excess electrons must be added to an isolated spherical conductor of diameer 30.0 cm to produce an electric field of 1155 N/C just outside the surface?

permittivity of free space and charge on an electron are given

2. A very long uniform line of charge has charge per unit length 4.62 \mu C/m and lies along the x-axis. A second long uniform line of charge has charge per unit length -2.46 \mu C/m and is parallel to the x-axis at y_1 = 0.414 m.

What is the magnitude of the net electric field at point y_2 = 0.188 m on the y-axis?

for this one i think it's along the lines of E = lamda/ 2 pi ep_0 r, but I'm not sure of which values are to be used


and...

3. The electric field at a distance 0.485 m from a very long uniform line of charge is 880 N/C.
How much charge is contained in a section of the line of length 2.00 cm?

i don't want the answer but help on how to solve the questions

 

[berkeman]

What equations relate charge distributions to electric field?

 

[kyle8921]

1. To find the number of excess electrons, we can use the formula for electric field: E = (1/4πε0)(q/r^2). Rearranging this formula, we get q = Er^2/(1/4πε0). Plugging in the given values, we get q = (1155 N/C)(0.15 m)^2/(1/4πε0) = 8.2 x 10^-7 C. Since each electron has a charge of -1.6 x 10^-19 C, we can divide the total charge by the charge of one electron to get the number of excess electrons, which is approximately 5.1 x 10^12 electrons.

2. To find the net electric field at point y_2, we can use the formula for electric field due to a line of charge: E = (λ/2πε0)(1/r). The electric field due to the first line of charge is E1 = (4.62 x 10^-6 C/m)(1/0.188 m) = 2.46 x 10^-5 N/C. The electric field due to the second line of charge is E2 = (-2.46 x 10^-6 C/m)(1/0.188 m) = -1.31 x 10^-5 N/C. Since the two lines of charge are parallel, the net electric field is simply the sum of these two fields, which is 1.15 x 10^-5 N/C.

3. The formula for electric field due to a line of charge can also be rearranged to find the charge: q = E(2πε0r). Plugging in the given values, we get q = (880 N/C)(2πε0)(0.485 m) = 2.7 x 10^-6 C. Since the length of the line is given in centimeters, we need to convert it to meters by dividing by 100. So the charge contained in a section of the line of length 2.00 cm is (2.7 x 10^-6 C)(0.02 m) = 5.4 x 10^-8 C.