A particle having charge q = +2.20 µC and mass m = 0.0100 kg is connected to a

Question

A particle having charge q = +2.20 µC and mass m = 0.0100 kg is connected to a string that is L = 1.20 m long and is tied to the pivot point P in the figure below. The particle, string, and pivot point all lie on a frictionless horizontal table. The particle is released from rest when the string makes an angle = 60.0° with a uniform electric field of magnitude E = 310 V/m. Determine the speed of the particle when the string is parallel to the electric field (point a in the figure).
_______m/s

Four identical point charges (q = +10.0 µC) are located on the corners of a rectangle, as shown in the figure below. The dimensions of the rectangle are L = 63.0 cm and W = 11.0 cm. Calculate the change in electric potential energy of the system as the charge at the lower left corner in the figure is brought to this position from infinitely far away. Assume that the other three charges in the figure below remain fixed in position.
____J

Explanation / Answer

(a)given that

q=2.20*10^(-6) C

m = 0.0100 kg

L = 1.20 m

E=310 V/m

theta =60 deg

this problem like an ideal pendulum in the gravitational field...
here m*g=q*E
so potential energy change =kinetic energy
(qE*L)-(qE*Lsin(theta)) =1/2*m*(v^2)

then v= sqrt((2*q*E*L*(1-sin(60))/m)

v=sqrt((2*(2.20*10^(-6))*310*1.20*(1-(-0.30)))/0.0100)

v=sqrt(0.113)

v=0.33 m/s

(b) given that L = 63.0 cm =0.63 m

W=11 cm = 0.11m

q1=q2=q3=q4=10*10^(-6) C

The system gains potential energy equal to the sum of the three potential energies between the introduced charge and the other three charges.
So it's going to be the sum of three different (1/4)*(q^2/r) expressions

one with r1 = 63 cm=0.63m

one with r2 = 11 cm=0.11 m and

one with r3^2 = ((0.63 m)^2 )+ ((0.11 m)^2)=0.3969+0.0121=0.409

r3=sqrt(0.409)

r3=0.64m

Technically, there are three more of these expressions (point charges at infinite distance), but they become zero.

so the total electric potential energy is

U=(k*q^2)*((1/r1^2)+(1/r2^2)+(1/r3^2))

U=(9*10^9)*(10*10^(-6))*((1/(0.63^2))+(1/(0.11^2))+(1/(0.64^2))

U=(90*10^3)*87.59 J

U=7.8*10^6 J

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