Two tiny 5.0-g ballooms are suspended from two very thin 1.0-m-long threads. The

Question

Two tiny 5.0-g ballooms are suspended from two very thin 1.0-m-long threads. The spheres repel each other after being charged to +91 nC and hang at rest as shown what is the angle theta? (k = 1/4pix0 = 8.99 times 10^9N middot m^2/C^2) 1. Charge Q is uniformly distributed around a thin conducting line of length 2a. Derive the electric field at a point P on the line axis at a distance x from its center. 2. The cross section of two spherical conductor shells is shown in the figure, with radii as given, and each shell is 2mm thick. The excess charge on the inner conductor is -30muC and the excess charges on the outer conductor is -70muC. The inner and outer surfaces are respectively denoted by A, B, C, and D, as shown A positive charge of 25muC is placed at the exact center of the spheres.

Explanation / Answer

Here,

m = 5 gm = 0.005 Kg

L = 1 m

q = 91 nC

distance between charges , d = 2 L * sin(theta)

d= 2 * sin(theta)

Now, electric force , F = k * q^2/d^2

balancing forces in vertical and horizontal direction

T * sin(theta) = F

T * cos(theta) = m * g

tan(theta) = F/(m * g)

tan(theta) = (k * q^2/d^2)/(m * g)

tan(theta) = (9*10^9 * (91 *10^-9)^2/(2 * sin(theta))^2)/(0.005 * 9.8)

solving for theta

theta = 4.51 degree

the angle theta is 4.51 degree

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