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# By how much does the spring stretch in cm?

Two 34*10^-6 C charges are attached to the opposite ends of a spring of spring constant 150 N/m and equilibrium length 50 cm .

### 2 Answers

- AvinashLv 71 decade agoFavorite Answer
Original length = 50 cm = 0.5 m

Let the spring stretch by distance x.

Then new length = 0.5 + x

Each charge Q = 34 * 10^-6 C

Spring constant k = 150 N/m

Electrostatic force = 9*10^9*Q^2/(0.5 + x)^2

Tension in the spring = (1/2)k*x^2

For equilibrium,

9*10^9*Q^2/(0.5 + x)^2 = (1/2)k*x^2

9*10^9* (34*10^-6)^2/(0.5 + x)^2 = (1/2)*150*x^2

10.404/(0.5+x)^2 = 75 * x^2

10.404/75 = x^2 * (0.5+x)^2

Taking square root on both sides,

sqrt(10.404/75) = x * (0.5 + x)

0.372 = x * (0.5 + x)

0.372 = 0.5 x + x^2

x^2 + 0.5 x - 0.372 = 0

x = [-0.5 +- sqrt(0.5^2 + 4*0.372)]/2

x cannot be negative.

x = [-0.5 + sqrt(1.738)]/2

x = (-0.5 + 1.32)/2

x = 0.82/2

x = 0.41 m

x = 41 cm

Ans: 41 cm