The figure shows an arrangement of four charged particles, with angle 0 = 33.0 d

Question

The figure shows an arrangement of four charged particles, with angle 0 = 33.0 degree and distance d = 2.00 cm. Particle 2 has charge q_2 = 9.60 Times 10^-19 C; particles 3 and 4 have charges q_3 = q_4 = -4.80 x 10^-19 C. (a) What is the distance D between the origin and particle 2 if the net electrostatic force on particle 1 due to the other particles is zero? (b) If particles 3 and 4 were moved closer to the x axis but maintained their symmetry about that axis, would the required value of D be greater than, less than, or the same as in part (a)? y (a) Number Units s s (b)

Explanation / Answer

here,

q2 = 9.6 * 10^-19 C

q3 = q4 = - 4.8 * 10^-19 C

theta = 33 degree

d = 2 cm

d = 0.02 m

let the force on charge 1 due to q3 be F

therefore due to symmetry,

the y component of the the force due to q3 and q4 cancels out

therefore , net force due to q3 and q4 = 2 * F * cos(theta)

for the net force on charge q1 to be zero

2 * F * cos(therta) = k * q1 * q2 / ( D + d)^2

2 * cos(33) * k * q1 * 4.8 * 10^-19 /0.02^2 = k * q1 * 9.6 * 10^-19 /( D + 0.02)^2

D = 1.84 * 10^-3 m

the distance D is 1.84 * 10^-3 m

b)

if particle 3 and 4 moves closer,

theta will decrease,

then the force due to 3 and 4 will increse and

so that to equal the forces,

the value of D also decreases

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