Hello I need solutions for those questions WITH steps plz. The electric field in

Question

Hello I need solutions for those questions WITH steps plz.

The electric field in a region of space has the components Ey = Ez =0 and Ex =(4.00 N/C)x. Point A is on the y axis at y 3.00 m: and point B is on the x axis at x=4.00 m. What is the potential difference VB ..VA? A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V = 0 at infinity). What is the radius of the drop?key: 5.4 times 104 m. If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop? Key: 22/3(500 V)= 790 V Figure shows a thin rod with a uniform charge density of 2.00 mC/m. Evaluate the electric potential at point P if d = D = L/4.00.

Explanation / Answer

dV = E*dr = -4 x dx

V = integral of - 4x dx from 0 to 4 = - 2 x^2 from 0 to 4 = -2*4^2 = -32 V

2) a) V= k q/r

r = k q/V = 9.0E9*3.0E-12/500= 5.4E-5 m

b)

so if V = 4/3 pi r^3

so if V double then r -> 2^(1/3) r

V = k 2 q/( 2^(2/3)r) = 2*500/2^(1/3)= 790 V

c)
consider a cut dx

r = sqrt( d^2 + x^2)

dV = k dq/r = k lambda dx/sqrt(d^2 + x^2) = 9.0E9*2.0E-3*dx/sqrt((L/4)^2 + x^2)

integrate both sides

V = 9.0E9*2.0E-6*integral of dx/sqrt((L/4)^2 + x^2) from L/4 to L

V=9.0E9*2.0E-6*4/L*integral of dx/sqrt( 1 + (4 x/L)^2) from L/4 to L

let u = 4x/L then du = 4 dx/L

V = 9.0E9*2.0E-6*integral of du/sqrt( 1 + (u)^2) from 1 to 4
= 9.0E9*2.0E-3*1.2133

V = 2.18E4 V

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