A small block of mass m and charge Q is placed on an...

A small block of mass $m$ and charge $Q$ is placed on an insulated, frictionless, inclined plane of angle $\theta$ as in Figure P23.51. An electric field is applied parallel to the incline. (a) Find an expression for the magnitude of the electric field that enables the block to remain at rest. (b) If $m=$ $5.40 \mathrm{g}, Q=-7.00 \mu \mathrm{C},$ and $\theta=$ $25.0^{\circ},$ determine the magnitude and the direction of the electric field that enables the block to remain at rest on the incline.

Key Concepts

Inclined Plane Analysis

The inclined plane setup involves breaking down the gravitational force into two components: one perpendicular to the plane and one parallel to it. The parallel component, given by mg sin?, is the component that causes motion along the plane. Understanding this decomposition is essential in solving problems involving forces on an inclined plane.

Electric Force on a Charged Particle

The force exerted on a charged object by an electric field is described by F = Q E, where Q is the charge and E is the electric field strength. The sign of the charge determines the direction of the force relative to the direction of the electric field. This concept is key when analyzing how an applied electric field can counteract other forces, such as the gravitational component along the incline.

Static Equilibrium

For any object to remain at rest, the net force acting on it must be zero. This condition of static equilibrium requires that all forces acting along a particular direction, such as along the incline, exactly balance each other. Establishing this balance allows one to calculate the necessary magnitude and direction of applied forces, like an electric force, to maintain a state of rest.

Force Balance Equations

Setting up force balance equations involves writing an equation where the sum of all forces acting along a given axis is zero. In this context, equating the gravitational component along the plane and the electric force along the same direction allows for the determination of the electric field needed to maintain equilibrium. This technique is a fundamental problem-solving tool in physics.

Transcript

00:01 Okay, so at the left corner, this is the graph i drew for this question.

00:05 So the block itself will experience gravity.

00:09 And if we decompose it, then you have two parts or two components.

00:16 Okay.

00:17 So if we consider x -axis is along the surface of the incline, and the y -axis is perpendicular to the incline surface.

00:25 So therefore, the x components of the gravity, which is m -g sine theta, is along the surface, and the y component is the mg cosine theta.

00:37 And so the electrofuels trying to pull the block at rest.

00:41 So if the sensors, as you can tell, the x component of the gravity is moving along the incline surface downward.

00:50 Okay? so in order to keep the block at rest, the electrow view must go against it, which is the opposite direction of the x components of the gravity.

00:59 So the magnitude of the electric force in this case should be equal to the magnitude of the x components of gravity.

01:09 So when the electric force is equal to charge times the electric view, which is qe.

01:12 So qe is the mg sine theta.

01:14 Therefore, e, which is the electric view, just simply equal to mg sine theta over q.

01:28 Okay.

01:30 And now for question b, these are the condition that was given from the question.

01:33 Okay...

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