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

Question

In a rectangular coordinate system a positive point charge 5.0 nC is placed at the point x = 2.0 cm, y = 0, and a point charge -5.0 nC is 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_e = 9.11 Times 10^-31 kg. elementary charge e = 1.60 Times 10^-19 C, k = 1/(4 pi epsion_0) = 8.99 Times 10 degree 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) due to 5nC

E1 = (kq/r^2) ( j)

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

due to - 5nC

angle by field vector with x axis, @ = 180 + tan^-1(3/4) = 217 deg

E2 = (9 x 10^9 x 5 x 10^-9 / (0.03^2 + 0.04^2)) (cos217i + sin217j)

E2 = -14375.44i - 10832.67 j N/C

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

magnitude = sqrt(14375.44^2 + 39167.33^2) = 41722.09 N/C

direction = 180 - tan^-1(39167.33 /14375.440 ) = 110.15 deg

b) V = [ kq1/r1] + [kq2/r2]

V= [(9 x 10^9 x 5 x 10^-9 / 0.03)] + [(9 x 10^9 x -5 x 10^-9)/sqrt(0.03^2 + 0.04^2)]

V = 600 Volt

c) using KE = - q deltaV

9.109 x10^-31 x v^2 /2 = 1.6 x 10^-19 x 600

v = 1.45 x 10^7 m/s

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