A charged cork ball of mass m is suspended on a light string...

A charged cork ball of mass m is suspended on a light string in the presence of a uniform electric field as shown in Figure $\mathrm{P} 23.59 .$ When $\overrightarrow{\mathbf{E}}=A \hat{\mathbf{i}}+B \hat{\mathbf{j}}$ , where $A$ and $B$ are positive numbers, the ball is in equilibrium at the angle $\theta .$ Find $(a)$ the charge on the ball and $(b)$ the tension in the string.

Key Concepts

Electrostatic Force

This represents the force experienced by a charged object in an electric field. The force is given by the product of the charge and the electric field vector and acts in the direction of the field for positive charges, while it reverses for negative charges. Understanding this concept is crucial for analyzing forces acting on charged bodies in uniform fields.

Gravitational Force

This is the force with which the Earth attracts any object with mass. It is given by the product of the object's mass and the acceleration due to gravity, acting in the downward direction. Recognizing the role of gravitational force enables the proper balancing of forces in problems involving suspended masses.

Static Equilibrium

An object in static equilibrium has zero net force acting on it. This means that all forces, including gravitational, electrostatic, and tension forces, balance each other out. This concept is essential to set up equations based on the equilibrium condition to solve for unknown quantities.

Force Decomposition

In problems involving forces at angles, it is necessary to break down vector forces into their orthogonal components, typically along horizontal and vertical axes. This technique simplifies the analysis by allowing separate consideration of forces in each direction, facilitating the application of equilibrium conditions.

Tension in a String

Tension is the force transmitted through a string or rope when it is pulled tight by forces acting from opposite ends. In equilibrium problems, the tension in the string balances both the weight of the object and any additional horizontal forces, making it key to solving the system of forces acting on the suspended mass.

Transcript

00:01 Okay, so this is the free body diagram for the charged cork ball.

00:04 Okay, so you should experience three different forces, which is gravity, electric force, due to the electric field, and the tension from the string.

00:11 So we can do some decomposing here.

00:12 So we can decompose the electric force and tension.

00:14 Therefore, we have the x component of electric force, which is f, e, x, and the y components of the electric force f, f, e .y.

00:23 And tension, we should have a x, y component as well, which is tx and ty.

00:29 And we know the angle here is theta.

00:31 So front question, the electric view was given as a vector quantity, which is ai plus b .j.

00:39 So the magnitude of the x components of the electric view should be equal to a.

00:42 And the magnitude of the magnitude of the y component of the electric field should be equal to b.

00:47 So therefore, we know the x components of the electric force should be equal to q times the magnitude of the s components of the electric field, which is qa.

00:56 And fey, which is the y components of the edge four, should be equal to qb.

01:00 Then we know the x component of tension is equal t -sign theta, okay? and the y components of retention is equal to t -cosin theta.

01:11 So, well, based on a graph, we know the x components of tension and the electrical force are having a same magnitude by opposite direction.

01:20 So therefore, they have same magnitude.

01:22 So therefore, we have t -x equal to f -e -x, which we have t -syn theta equal to q -a.

01:28 And we also know that the y -component.

01:32 Of the tension plus the white components of the electrical force is equal to the gravity.

01:37 Okay? so therefore we have the wide components of tension is equal to gravity mg minus f, b...

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