Question
I'm not sure if we have learned this or not, so please give a through explanation.
The electric field 22 cm from a long wire carrying a uniform line charge density is 1.7 kN/C. What will be the field strength 37 cm from the wire?
What is the line charge density on a long wire if the electric field 48 cm from the wire has magnitude 250 kN/C and points toward the wire?
How strong an electric field is needed to accelerate electrons in a TV tube from rest to one-fourth the speed of light in a distance of 7.0 cm?
I know it's three questions, but I think they are all short if you know how to do them, please help
Explanation / Answer
Given: (a) electric field of long wire carrying a uniform line charge E = 1.7 k N / C = 1700 N / C At a distance : r = 22 cm = 0.22 m It is known by the formula for Electri field due to Line of charge E = / 2o r where is the charge density of the wire thus, = E (2o r) = (1700)( 2o x 0.22) = 374 ( 2o ) C/ m for this case , E = ? when , r = 37cm = 0.37 m so, E = / 2o r = 374 ( 2o ) / (2o ) r = 374 / 0.37 = 1010.81 N / C (b) point distnce : r = 48 cm = 0.48 m electric field from the wire has magnitude: E = 250 kN/C = 250000 N / C = ? so, E = / 2o r = E ( 2o ) r = E (4 o ) r /2 but , 1/ 4 o = 9 x 109 Nm2/ C2 = (250000 x 0.48) / (2x 9x109) = 6666.7 x 10-9 C /m (c ) Intial velocity of the electrons : U = 0 m/s final velocity : V = c /4 = 0.75 x108 m/s distance traveled : S = 7 cm = 0.07 m Let , a be the acceleration attained in the electron so, V2 - U2 = 2 aS a = V2 / 2 S = ( 0.75 x108 )2 / 2 (0.07) = 4.017 x 1016 m/s2 From , Newton's law F = q E = ma E = ma / q = (9.1x10-31)(4.017 x 1016 ) / (1.6x10-19) = 228466.8 N / C (b) point distnce : r = 48 cm = 0.48 m electric field from the wire has magnitude: E = 250 kN/C = 250000 N / C = ? so, E = / 2o r = E ( 2o ) r = E (4 o ) r /2 but , 1/ 4 o = 9 x 109 Nm2/ C2 = (250000 x 0.48) / (2x 9x109) = 6666.7 x 10-9 C /m (c ) Intial velocity of the electrons : U = 0 m/s final velocity : V = c /4 = 0.75 x108 m/s distance traveled : S = 7 cm = 0.07 m Let , a be the acceleration attained in the electron so, V2 - U2 = 2 aS a = V2 / 2 S = ( 0.75 x108 )2 / 2 (0.07) = 4.017 x 1016 m/s2 From , Newton's law F = q E = ma E = ma / q = (9.1x10-31)(4.017 x 1016 ) / (1.6x10-19) = 228466.8 N / C