The figure shows an arrangement of four charged particles, with angle = 40.0 and

Question

The figure shows an arrangement of four charged particles, with angle = 40.0 and distance d = 3.00 cm. Particle 2 has charge q2 = 8.00 × 10-19 C; particles 3 and 4 have charges q3 = q4 = -3.20 × 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)?

https://ibb.co/knvHwk

Explanation / Answer

(a) Let us say that charge on particle 1 is q,

then the force on it from particle 2 is 8kq/(d+D)2, where k=1/40.

The y-components of forces on particle 1 from charges 3 and 4 are equal and opposite, so we can cancel the y - components;

the x-components are each -1.6kqcos/r2 where r is the distance between 1 and 4.

We have rcos = d and cos = 0.77, so (cos)2= 0.593

cos/r2 = cos x (cos)2/d2 = 0.77 x 0.5932 / d2 = 0.271/ d2 .

sum of x-components of forces on 1 from charges 2, 3 and 4 should be 0.
8*kq/(d+D)2 - 2 x 3.2 kq* 0.271/ d2= 0. ------- (1)

Insert d= 3 cm in equation (1) We get,
(D+3)2 = 4*9 / (3.2*0.271) =41.51

or D+3 = 6.44

Therefore D = 6.44-3 =3.44 cm

(b) Since the angle decreases the cosine value increases, so contribution from y-component increases, to offset this force exerted by q2 should be stronger to bring it closer to q1, so D must be decreased or less than same as part (a)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.