two charges +q and 2q are held in the place on the x axis and the locations x=-d

Question

two charges +q and 2q are held in the place on the x axis and the locations x=-d and x=+d respectively.

a) find the magnitude and the direction of the electric field at the location indicated in the figure below. use q=1x10^-6 and d=0.1m.

b) find the voltage at two charges at location a

c) find the voltage at two charges at location b

d) A-5 x10-6 charge is released from rest at the point Find its velocity if it crosses the point at a later time.Themass of the charge is110?6kg.

two charges +q and 2q are held in the place on the x axis and the locations x=-d and x=+d respectively. a) find the magnitude and the direction of the electric field at the location indicated in the figure below. use q=1x10^-6 and d=0.1m. b) find the voltage at two charges at location a c) find the voltage at two charges at location b d) A-5 x10-6 charge is released from rest at the point Find its velocity if it crosses the point at a later time.Themass of the charge is1½10?6kg. .e) A charge of 10x10-6 C is placed at a. Find the magnitude and direction of the force exerted on this charge

Explanation / Answer

so we will add by components

so Ex = Ex from q + Ex from 3q

= k q/(d^2 + d^2) cos(45) - k 3q/(d^2+d^2) cos(45)

= - 2 k q/(2 d^2) cos(45) = -k q/d^2 cos(45) = -9.0E9*1.0E-6/0.1^2*cos(45)

Ex = -6.36E5 N/C

now y direction

Ey = k q/(d^2 + d^2) sin(45) + k 3q/(d^2+d^2) sin(45)

= 4 k q/(2 d^2) sin(45) = 2 k q/d^2 sin(45) = 2*9.0E9*1.0E-6/0.1^2*cos(45)

=1.27E6 N/C

so magnitude = sqrt(Ex^2 + Ey^2) = sqrt(6.36E5^2 + 1.27E6^2)= 1.42E6 N/C

direction = arctan(y/x) = arctan(1.27E6/-6.36E5) = 116.6 degrees

b) V = sum of kq/r = k q/r + 3 kq/r = 4 kq/r = 4*9.0E9*1.0E-6/(sqrt(0.1^2 + 0.1^2))= 2.55E5

c) now r = d

V = 4*9.0E9*1.0E-6/(sqrt(0.1^2 + 0.1^2))= 3.6E5

d) q V = q V + 1/2 mv^2

-5.0E-6*2.55E5 = -5.0E-6*3.6E5 + 0.5*1.0E-6*v^2

v=1025 m/s

e) F = q E= 10.0E-6*1.42E6= 14.2 N

same direction 116.6 degrees

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