1) The drawing shows six point charges arranged in a rectangle. The value of q i

Question

1) The drawing shows six point charges arranged in a rectangle. The value of q is 1.65 ?C, and the distance d is 0.253 m. Find the total electric potential at location P, which is at the center of the rectangle.

2) Two identical point charges (q = +5.70 x 10-6 C) are fixed at opposite corners of a square whose sides have a length of 0.560 m. A test charge (q0 = -4.20 x 10-8 C), with a mass of 7.00 x 10-8 kg, is released from rest at one of the corners of the square. Determine the speed of the test charge when it reaches the center of the square.

Explanation / Answer

a)

Potential at a point due to a charge at distance r is given by:

V = kq/r

where , k = 9*10^9

q = charge

r = distance of the point from the charge

So,

potential due to charge +7q at the top left corner , V1 = k*(7q)/(sqrt(5)*d/2)

here r = distance of the center of the rectange from the left vertex = sqrt(5)*d/2 <----this can be found by simple trigonometry(by calulating the half distance of the length of the diagonal of the rectange)

Similarly for the +3q charge at the top, V2 = k(3q)/(d/2)

for +5q at the top right, V3 = k(5q)/(sqrt(5)*d/2)

for +7q charge at bottom right, V4 = k(7q)/(sqrt(5)*d/2)

for -3q charge at bottom, V5 = k(-3q)/(d/2)

for -5q, V6 = k(-5q)/(sqrt(5)*d/2)

So, net potential, V = V1+V2+V3+V4+V5+V6 = 2*k*(7q)/(sqrt(5)*d/2

= 2*(9*10^9)*(7*1.65*10^-6)/(sqrt(5)*0.253/2)

= 7.35*10^5 V <-------------answer

b)

By conservation of energy, K.E gained = change of electric potential energy

Change of Potential energy(P.E) = P.E at center - P.E at A

By similar method as in part (a), we can calulate the potential at A, Va = 2*kq/d = 2*(9*10^9)*(5.7*10^-6)/(0.56)

= 1.83*10^5 V

So, potential energy at A, Ua = Va*qo = 1.83*10^5*(4.2*10^-8) = 7.69*10^-3 J

Similarly potential energy at center, Uc = 2*k(q*qo)/(d/sqrt(2)) = 2*9*10^9*(5.7*10^-6*4.2*10^-8)/(0.56/sqrt(2)

So, Uc = 1.1*10^-2 J

So, change in P.E = Uc - Ua = 1.1*10^-2 - 7.69*10^-3 = 3.31*10^-3 J

So, By conservation of energy, K.E = change in P.E

So, 0.5*m*v^2 = 3.31*10^-3

So, v = sqrt(2*3.31*10^-3/m) = sqrt(2*3.31*10^-3/(7*10^-8)) = 307.5 m/s <------------------answer

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