An electric field of magnitude14000 N/C and directed upward perpendicular tothe

Question

An electric field of magnitude14000 N/C and

directed upward perpendicular tothe Earth’s

surface exists on a day when athunderstorm

is brewing. A truck that can beapproximated

as a rectangle3.2m by 2.3 m is traveling

along a road that is inclined11? relativeto

the ground.

Determine the electric fluxthrough the

bottom of the truck. Answer inunits of

Through whatpotential difference would an

electron needto be accelerated for it to

achieve aspeed of 5.3% of the speed of light

(2.99792 × 108 m/s), starting from rest?An-

swer in units ofV.

Explanation / Answer

Electric field E = 14000 N / C Area A = 3.2 m * 2.3 m             = 7.36m ^ 2 angle = 11 degrees required flux = EA cos                     = 101146.86 N m^ 2 / C (b). Speed v = 5.3 % of 2.99792* 10 ^ 8 m / s                     = 0.15888* 10 ^ 8 m / s required potential difference V = ( 1/ 2q ) m v ^2                                                 =[1/ ( 2* 1.6 * 10 ^ -19 ) ] * ( 9.11 * 10 ^ -31 ) ( 0.15888* 10 ^ 8) ^ 2                                                = 0.07186 * 10 ^ 4 m / s (b). Speed v = 5.3 % of 2.99792* 10 ^ 8 m / s                     = 0.15888* 10 ^ 8 m / s required potential difference V = ( 1/ 2q ) m v ^2                                                 =[1/ ( 2* 1.6 * 10 ^ -19 ) ] * ( 9.11 * 10 ^ -31 ) ( 0.15888* 10 ^ 8) ^ 2                                                = 0.07186 * 10 ^ 4 m / s

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