Question

Two thin hollow plastic spheres, about the size of a ping-pong ball with masses ( 0.004 kg) have been rubbed with wool. Sphere 1 has a charge -4 × 10-9 C and is at location < 0.47, -0.29, 0 > m. Sphere 2 has a charge -9 × 10-9 C and is at location < -0.42, 0.43, 0> m. It will be useful to draw a diagram of the situation, including the relevant vectors.

(a) What is the relative position vector pointing from q1 to q2?

(b) What is the distance between q1 and q2?

(c) What is the unit vector r^ in the direction of r→?

(d) What is the magnitude of the gravitational force exerted on q2 by q1?

(e) What is the gravitational force (vector) exerted on q2 by q1?

(f) What is the value of (1/4πɛ0)*(q1q2/r²)?

(g) What is the electric force (vector) exerted on q2 by q1?

(h) What is the ratio of the magnitude of the electric force to the magnitude of the gravitational force?

(i) If the two masses were 5 times farther away (that is, if the distance between the masses were 5 ×|r→|), what would be the ratio of the magnitude of the electric force to the magnitude of the gravitational force now?

Answer 1

The position vector for sphere 1 is (0.47 mi-(0.29 m The position vector for sphere 2 is: ,--(0.42 m)? +(0.43 m) The position vector from q,to gjis 122 (-(0 0.42 mi +(0.43 m)j -(047 m)i -(0.29 m)J) -(0.89 m)i +(0.72 m)j The distance can be calculated by calculating the magnitude of position vector 0.89 m)i +(0.72 m)j İ2 (-0.89 m)' + (0.72 m) 2-1.14nm

The unit vector is: 12 1. Fa 12 -(0.89 m)i +(0.72 m)j 1.1Am i-(0.78)i +(0.63)J The gravitational force vector when force is exerted onq, byq, '12 (i2) The unit position vector: 21 12 Gravitational force exerted onq, by q, is

21 -2 21 21 (%)12 6.67x1011 N-m' kg2) (0.004 kg 6108)+(063)) 21 (1.14 m) = 8.21x 10-16 N((0.78)^-(0.63) F, -(64x10* N-(517x10- N