Electrostatics question?

A spring having a spring constant of 4.8 N / m and a length of 0.046 m when at rest is suspended vertically from a support. A uniform electric field, directed upward, surrounds the spring. A sphere holding a charge of +7.1 μC and having a mass of 0.0052 kg is attached to the lower end of the spring, and the spring-sphere system is allowed to reach equilibrium wherein the length of the spring becomes 0.033 m. Determine the magnitude of the electric field. A) 2.8 x 103 N / C B) 4.5 x 103 N / C C) 7.9 x 103 N / C D) 1.6 x 104 N / C E) 4.1 x 104 N / C

Jun 1, 2018

$\text{Option D is true.}$

Explanation:

• When the sphere is connected to the spring end, the length of the spring must be increased. We see that the length of the spring is decreasing.

0.046-0.033 = 0.013 meters.

• Electrostatic force must have exerted upward on the ball bearing the electric charge. Positive electric charges affect the same direction as the direction of the electric field. The direction of the electric field is straight up.

• We can calculate the net force acting on the spring using the formula F = $K \cdot \Delta x$
In this formula, K is the spring constant , $\Delta$x is the amount of shortening.

• F_("net")=4.8 *0.013=0.0624 " "N

• Since the weight vector is downwards, the electrostatic force vector is upward, the magnitude of the net force is equal to the difference of the two vectors.

• ${F}_{\text{net}} = q E - m g$

• Where qE is the electrostatic force acting on the sphere, mg is the weight of the sphere. $0.0624 = 7.1 \cdot {10}^{-} 6. E - 0.0052 \cdot 9.81$

• $0.0624 = 7.1 \cdot {10}^{-} 6. E - 0.0510$
$0.0624 + 0.05101 = 7.1 \cdot {10}^{-} 6. E$
$0.11341 = 7.1 \cdot {10}^{-} 6. E$
$E = \frac{113410}{7.1} =$

$E = 1.6 \cdot {10}^{4} \frac{N}{C}$