In a rectangular coordinate system a positive point charge 5.0nC is placed at th

Question

In a rectangular coordinate system a positive point charge 5.0nC is placed at the point x - 2.0 cm, y - 0, and a point charge 5.0 nC placed at x = -2.0 cm, y = 0. Point P is at x = 2.0 cm, y -3.0 cm. The electric potential at infinity is zero. Mass of electron m_c - 9.11 Times 10^31 kg, elementary charge c - 1.60 Times 10^-19 C, k = 1/(4xt_0) = 8.99 Times 10^9 N-m^2/C^2. Find the magnitude and direction of the electric field at point P. Find the electric potential at point P. An electron, initially at rest at a point midway between the two charges, travels to point P. What is its speed at point P?

Explanation / Answer

a) magnitude of field strength due to a point charge at distanc r = kq / r^2

due to 5nC :

r = 3 cm

point P is just above the charge so field due to +ve charge is directly away from charge.

(hence along j )

E1 = (9 x 10^9 x 5 x 10^-9) / (0.03^2) = 50000 N/C

due to -5nC :

r = sqrt((2+2)^2 + 3^2 ) = 5 cm

towards the negative charge.

angle = 180 + tan^-1(3/4) = 217 deg

E2 = [(9x10^9 x 5 x 10^-9)/0.05^2] ( cos217i + sin217j)

E2 = - 14375.44i - 10832.67j N/C

Enet =E1 +E2 = - 14375.44i + 39167.33j N/C

magnitude = sqrt(Ei^2 + Ej^2) = 41722.09 N/C

direction = 180 - tan^-1(39167.33 / 14375.44) = 110.15 deg

b) PE = kq / r

Vnet = [ (9 x 10^9 x 5 x 10^-9) / 0.03 ] + [ (9 x 10^9 x -5 x 10^-9 ) / 0.05 ]

= 600 volt

c) point at midway will be at same distance from each charge. (2cm from each)

so Vnet = [ (9 x 10^9 x 5 x 10^-9) / 0.02 ] + [ (9 x 10^9 x -5 x 10^-9 ) / 0.02 ]

= 0

Energy gain = q deltaV

m v^2 / 2 = q (0 - 600 )

9.109 x 10^-31 x v^2 / 2   = -1.6 x 10^-19 x -600

v = 1.58 x 10^7 m/s

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