A charged cork ball of mass 0.60 g is suspended on a light string in the presence of a uniform electric field as shown in the figure below. When E = (4.30 i +4.10 j) x 105 N/C, the ball is in equilibrium at 0 = 37.0⁰. E (a) Find the charge on the ball. (b) Find the tension in the string. @
A charged cork ball of mass 0.60 g is suspended on a light string in the presence of a uniform electric field as shown in the figure below. When E = (4.30 i +4.10 j) x 105 N/C, the ball is in equilibrium at 0 = 37.0⁰. E (a) Find the charge on the ball. (b) Find the tension in the string. @
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#### (a) Find the charge on the ball.
#### (b) Find the tension in the string.
### Part 1 of 4 - Conceptualize
The electric force is in the same direction as \(\vec{E}\) since the ball is positively charged. If we examine the free body diagram that shows the three forces acting on the ball, the sum of which must be zero, we can see that the tension in the string is about half the magnitude of the weight of the ball.
**Free Body Diagram:**
1. Tension in the string \(T\) acting upward at an angle \(\theta\).
2. Weight \(mg\) acting downward.
3. Electric force \(q\vec{E}\) acting in the direction of the electric field.
### Part 2 of 4 - Categorize
The tension in the string is found by applying Newton's second law to this static (electrostatic, in this case) problem. Since the force vectors are in two dimensions, we must consider both the \(x\) and \(y\) components of \(\sum \vec{F} = m\vec{a}\).
### Part 3 of 4 - Analyze
From Newton's second law, we know that the sum of the forces on the ball is equal to zero. We have:
\[
\sum \vec{F} = \vec{T} + q\vec{E} + \vec{F}_g = 0
\]
For equilibrium in the \(x\) and \(y\) directions, we have the following equations.
\[
\sum F_x = qE_x - T](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc372217a-7cd5-47a2-b963-c57d0a9e3896%2F421de3f3-0220-4674-8a16-180943e35f79%2F9zm1ex_processed.png&w=3840&q=75)
Transcribed Image Text:### Charged Cork Ball in Uniform Electric Field
A charged cork ball of mass \(0.60 \, \text{g}\) is suspended on a light string in the presence of a uniform electric field as shown in the figure below. When \(\vec{E} = (4.30 \, \hat{i} + 4.10 \, \hat{j}) \times 10^3 \, \text{N/C}\), the ball is in equilibrium at \(\theta = 37.0^\circ\).
#### (a) Find the charge on the ball.
#### (b) Find the tension in the string.
### Part 1 of 4 - Conceptualize
The electric force is in the same direction as \(\vec{E}\) since the ball is positively charged. If we examine the free body diagram that shows the three forces acting on the ball, the sum of which must be zero, we can see that the tension in the string is about half the magnitude of the weight of the ball.
**Free Body Diagram:**
1. Tension in the string \(T\) acting upward at an angle \(\theta\).
2. Weight \(mg\) acting downward.
3. Electric force \(q\vec{E}\) acting in the direction of the electric field.
### Part 2 of 4 - Categorize
The tension in the string is found by applying Newton's second law to this static (electrostatic, in this case) problem. Since the force vectors are in two dimensions, we must consider both the \(x\) and \(y\) components of \(\sum \vec{F} = m\vec{a}\).
### Part 3 of 4 - Analyze
From Newton's second law, we know that the sum of the forces on the ball is equal to zero. We have:
\[
\sum \vec{F} = \vec{T} + q\vec{E} + \vec{F}_g = 0
\]
For equilibrium in the \(x\) and \(y\) directions, we have the following equations.
\[
\sum F_x = qE_x - T
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