(a) In Figure 24-33, what is the net electric potential (voltage) at point p due

Question

(a) In Figure 24-33, what is the net electric potential (voltage) at point p due to the four fixed particles if V = 0 at infinity, q = 655 nC, and d = 1.50 cm? t9 Fig. 24-33 (b) Suppose we move a fifth particle that has charge 226 nC starting from point P and ending up at infinity moving to the left along a line extending through the two negative charges. What would be the change in the electric potential energy for this fifth particle? (c) If a different path were taken between P and infinity, the potential energy change would be smaller O would be larger O could be larger or smaller depending on the path. would be the same

Explanation / Answer

Given

charge q = 655 nC , d = 0.015 m

and we know the potential is V = kq/r

here the potential at point P due the charges is

V (p) = kq (1/d+1/d -1/d-1/2d)

V(p) = kq(1/d)(1-1/2)

v(p) = kq(1/2)(1/d)

v(p) = kq/2d

substituting the values

v(p) = (9*10^9*655*10^-9)/(2*0.015) V

v(p) = 196500 V

b) if the fifth charge q5 = 226 nC is at P the potential energy is  

U = kq1*q2/r^2

U1 = kq*q5(1/d^2+1/d^2 -1/d^2-1/4d^2)

U1 = kq*q5*1/d^2(1-1/4)

U1 = 3/4*k*q*q5/d^2

U1 = (9*10^9*655*10^-9*226*10^-9)(3/(4*0.015^2)) J

U1 = 4.4409 J

at infinity the potential energy is zero  

so the potential energy difference is 4.4409 J only

c) no change , would be the same  

----------------

v(r) = kq/r

a) q = -5.0 nC at origin , at x= 0.5 m

the potential is V(0.5) = 9*10^9*(-5*10^-9)/(0.5) V = -90 V

b) at point (x,y) = (0.3m,0.4 m)

the distance from origin to the point is r = sqrt(0.3^2+0.4^2) = 0.5 m

so potential at 0.5 m is v(0.3m,0.4m) = 9*10^9*(-5*10^-9)/(0.5) V = -90 V

c) second charge q2 = 11 nc placed at x = -1.1 m then what is the potential at x = 0.5 m ?

the potential is V2 = 9*10^9*(11*10^-9)/(1.1+0.5) v = 61.875 V

net potential is at x=0.5 due to both charges is  

V = -90+61.875 V = -28.125 V

d) at y = 1.9 m

due to first charge v1 = 9*10^9*(-5*10^-9)/(1.9) V= -23.68 V

DUE TO second charge v2 = 9*10^9*(11*10^-9)/(sqrt(1.1^2+1.9^2)) V = 45.1 V

net potential is V = v1+v2 = -23.68+45.1 V = 21.42 V

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