Question

An unstrained horizontal spring has a length of 0.29 m and a spring constant of 180 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 0.021 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.

Answer 1

A) they both have the same algebraic sign

B)6.377×10^-6 C

Step-by-step explanation:

From columb's law, the force acting on both charges can be expressed as

F=( kq1*q2)/r^2

Where F= electrostatic force

r= distance between the charges

q1 and q2= charges

The force acting on a spring can be expressed as

F= kx..................eqn(2)

Where

K= spring constant = 180 N/m.

x= stretch of the string= 0.021m

Substitute the values into eqn (2)

F= (180×0.021)

F= 3.78N

If we compare with spring force,

Hence, F( electrostatic) = 3.78N

From

F=( kq1*q2)/r^2 ..............eqn(1)

Where

r= (0.29 m + 0.021m)= 0.311m

K= the electrostatic constant= 8.99×10^9 kg⋅m3⋅s−2⋅C−2.

If we substitute the values we have

Since the charges are the same, then

kq1 and q2 equals "q"

3.78= (8.99×10^9 ×q^2)/(0.311)^2

Making q^2 subject of the formula

3.78× (0.311)^2 = 8.99×10^9 × q^2

q^2= [(0.311)^2 × 3.78]/8.99×10^9

q^2= 40.668×10^-12

q=√40.668×10^-12

q= 6.377×10^-6 C

(a) the possible algebraic signs

They have the same algebraic sign

(b) the magnitude of the charges.

6.377×10^-6 C