Question

Part A
Find the electric potential at point P in the figure

Vp=________V

Part B

Suppose the three charges shown in the figure are held in place. A fourth charge, with a charge of +7.00 and a mass of 4.80 , is released from rest at point P. What is the speed of the fourth charge when ithas moved infinitely far away from the other three charges?

v=________m/s

Answer 1

Part A :

AP = SQRT(AC² - AP²) = ( 1.5625 - .3906) = 1.1719m
using k = 4πε₀
potential at P due to charge at A = V1 = k * 7.45 * 10^-6 / r1 [r1 =1.17m]
= 57.30 * 10³V

potential at P due to charge B = V2 = k * 2.75 * 10^-6 /r2 [r2 = 0.625m]
= 39.6 * 10³ V


potential at P due to charge C = V3= k * (-1.72 * 10^-6) / r3[r3 =0.625m]
= -24.768 * 10³ V

potential at P = V_p = V1 + V2 + V3
= (57.30 + 39.6 - 24.768) * 10³ V
= 72.132 * 10³ V

Part B :
now if charge of 7μC is placed at P potential energy in the system because of charge at P will be
=potential energy due to combination of (7.45μC and 7μC) + potential energy due to combination of (7.45μC and 2.75μC) + potential energy due tocombination of (7.45μC and -1.72μC)
=7.45 * V1 + 7.45 * V2 + 7.45 * V3
=7.45 * 57.30 * 10³ V + 7.45 * 39.6 * 10³ V - 7.45 * 24.768 * 10³ V
=537.3834 * 10³ J

there is no force on the charge at infinity due to any of the charge of triangle
the energy in the system which is due to charge at P will be converted to kinetic energy of that charge at infinity

using CONSERVATION OF ENERGY PRINCIPLE

=> 537.38 * 10³ = 1/2 M V²
=> 537.38 * 10³ = 1/2 * 4.80 * 10^-3 * V²
=> V² = 537.38 * 2 / 4.80
=> V² = 223.90
=> V = 14.96 m/s

so this should be the answer according to the cocept used by me here.

Answer 2

electric potential = k*q/r

so potential at P due to +2.5µC charge is which is at distance = 1.25/2 = 0.625 mts is

= (9*10^9)*(2.5µC) / (0.625) = 36*10^3 N-m/C

so potential at P due to -1.72µC charge is which is at distance = 1.25/2 = 0.625 mts is

= (9*10^9)*(-1.72µC) / (0.625) = -24768 N-m/C

so potential at P due to 7.45µC charge is which is at distance = altitude of equilateral triangle of side 1.25 mts

altitude = a(sqrt(3)/2) = 1.08253175 mts

= (9*10^9)*(7.45µC) / (1.08253175) = 61938.1371 N-m/C

so net potential is sum of these potentials =

61938.1371 + 36000-24768 = 73170.1371 N-m/C


b) if a charge of 7 µC is kept at P

then potential energy = charge * pontential at P

= (7µC)*(73170.1371 ) = 0.512 19096 N-m


when it moves to infinity this energy is converted to kinetic energy


so (1/2)*m*(v^2) = 0.51219096

m = 4.8 gms = 0.0048 kgs

V^2 = (2*0.51219096)/0.0048 = 213.4129

V = sqrt(213.4129) = 14.6086584 m/s

velocity it acheives = 14.6086584 m/s

Answer 3

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