What is the charge on one of the balls?---**Question Stem:**
As shown to the right, two identical 1.0 x 10⁻⁴ kg balls carry identical charges and are suspended by two threads of equal length. Find the charge on either ball. (Hint: examine one ball only; the F_E on it is the same as for the other)
**Diagram Description:**
* **Type:** Physics diagram showing two identical objects suspended by threads.
* **Main Elements:**
* Two black circular balls.
* Two straight lines representing threads, suspending the balls from a horizontal line representing a support. The threads meet at a point on the support.
* A horizontal line at the top representing the support.
* An angle of 60 degrees is labeled between the vertical direction (indicated by a dashed line pointing upwards towards the suspension point) and the thread on the left side, adjacent to the left ball.
* An angle of 60 degrees is labeled between the vertical direction (indicated by a dashed line pointing upwards towards the suspension point) and the thread on the right side, adjacent to the right ball.
* A horizontal dotted line connects the centers of the two balls.
* A double-headed arrow below the dotted line indicates the distance between the centers of the two balls.
* The distance between the centers of the two balls is labeled as 40 cm.
* **Relative Position and Direction:** The two balls are suspended below the support, separated horizontally. The threads are angled outwards from the vertical. The 60 degree angles are measured from the vertical. The distance between the balls is horizontal.
**Other Relevant Text:**
* Mass of each ball: 1.0 x 10⁻⁴ kg
* Charges: Identical
* Threads: Equal length
* Hint: examine one ball only; the F_E on it is the same as for the other (F_E likely represents the electrostatic force).
* Angle of thread with vertical: 60 degrees
* Distance between balls: 40 cm
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答案文本
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We have two identical charged balls suspended by threads. Each ball has a mass of 1.0 times 10 to the minus 4 kilograms. The threads make 60-degree angles with the vertical, and the balls are separated by 40 centimeters. Since the balls repel each other, they must have the same type of charge. Our goal is to find the magnitude of the charge on each ball.
To solve this problem, we analyze the forces acting on one ball. There are three forces: the weight W equals mg acting downward, the tension T along the thread, and the electrostatic force F-E acting horizontally away from the other ball. We can resolve the tension into vertical and horizontal components. The vertical component is T cosine 60 degrees, and the horizontal component is T sine 60 degrees.
For equilibrium, the net force in each direction must be zero. Vertically, T cosine 60 degrees equals W. Horizontally, T sine 60 degrees equals F-E. Dividing these equations gives us tangent 60 degrees equals F-E over W. Therefore, F-E equals W times tangent 60 degrees. The weight W equals mg, which is 1.0 times 10 to the minus 4 times 9.8, giving us 9.8 times 10 to the minus 4 Newtons. The electrostatic force is then 9.8 times 10 to the minus 4 times square root of 3, which equals 1.697 times 10 to the minus 3 Newtons.
Now we apply Coulomb's Law. The electrostatic force equals k times the product of charges divided by distance squared. Since both charges are identical, this becomes k times q squared over r squared. Substituting our values: 1.697 times 10 to the minus 3 equals 9 times 10 to the 9 times q squared over 0.40 squared. Solving for q squared gives us 3.017 times 10 to the minus 14. Taking the square root, the magnitude of the charge is 1.74 times 10 to the minus 7 Coulombs.
We have successfully solved the problem. The magnitude of the charge on each ball is 1.74 times 10 to the minus 7 Coulombs. Since the balls repel each other, they must have charges of the same sign, either both positive or both negative. Our solution involved analyzing the equilibrium forces, finding the electrostatic force using trigonometry, and then applying Coulomb's Law to determine the charge magnitude.