An unstrained horizontal spring has a length of 0.43 m and a spring constant of 238 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.013 m relative to its unstrained length. Determine the possible algebraic signs and the magnitude of the charges. (a) the possible algebraic signs

The charges on the spring are 1.23E-5 C and they have the same sign

Explanation:

Given

Coulomb laws states that the force exerted by a charge q on another charge Q at a distance r is given by

F = kqQ/r²

Where k = 8.99 * 10^9 Nm²/C²

r = 0.43 + 0.013

r = 0.443m

The force on the spring is calculated as;.

F = kx where x is the stretch length of the spring and k is the spring constant

The force acting on the spring = 238 * 0.013

F = 3.094N

By comparison;

F = kqQ/r² becomes

3.094 = F = kqQ/r²

kqQ/r² = 3.094 (Considering that a = Q)

kq²/r² = 3.094

8.99 * 10^9 * q²/0.443 = 3.094 -- make q² the subject of formula

q² = 3.094 * 0.443/8.99*10^9

q² = 1.524629588431E−10

q = √1.524629588431E−10

q = 0.000012347589191545C

q = 1.23E-5 C

The charges on the spring are 1.23E-5 C and they have the same sign

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.