A displacement vector points in a direction of θ = 23° left of the positive y-axis. The magnitude of this vector is D = 155 m. R

Answer: A) 2 B) 4 C) 1

Explanation:

The electric field associated with a parallel-plate capacitor is defined as:

A) E=Q/(L^2 * ε0); if we double the charge, the resultant electric field becomes twice as strong as its initial value.

B) Referring to the earlier electric field expression, if the plate size is doubled, the final electric field will be a quarter of the original strength.

C) If the separation distance between the plates is increased twofold, the resulting electric field remains unchanged from its initial state.

According to Newton's second law, provided F is directed horizontally,

• the net horizontal force acting on the larger block is

F - µmg = 3mA

where µmg represents the friction experienced by the larger block from contact with the smaller block, µ is the static friction coefficient between both blocks, and A indicates the acceleration of the block;

• the net vertical force on the larger block is

4mg - 3mg - mg = 0

where 4mg denotes the magnitude of the normal force exerted by the surface on the combined mass of both blocks, 3mg corresponds to the weight of the larger block, and mg indicates the weight of the smaller block;

• the net horizontal force acting on the smaller block is

µmg = ma

where µmg again signifies the friction between both blocks; however, it is important to note that this force aligns in the same direction as F. It is the sole force influencing the smaller block in the horizontal direction, thus (b) static friction causes the smaller block's acceleration;

• the net vertical force on the smaller block is

mg - mg = 0

where mg represents the force of both the normal force from the larger block pushing up against the smaller one, and the weight of the smaller block.

(You should be able to create your own free-body diagrams based on the forces discussed.)

(c) Solve the equations stated above to find A and a:

A = (F - µmg) / (3m)

a = µg

The electric flux through the cylindrical surface surrounding the infinite charged wire is given by the formula ∅E = E x 2πrl. To analyze this, we consider an infinitely long straight wire with a uniform linear charge density of λ Cm⁻¹. The electric field at a distance r from this charge can be evaluated using a cylindrical Gaussian surface of radius r and length l, oriented along the wire. Only the curved surface of the cylinder contributes to the total flux since the other surfaces are perpendicular to E.

Answer:

The potential energy tied to the charge diminishes.

The electric field performs negative work on the charge.

Explanation:

The resulting motion can be determined using the Pythagorean theorem, as the two components (north and east) are at right angles. To find the direction, trigonometry is applied, yielding Ф=arctan(3.8/12)=17.57° north of east.