Recall that there is a uniform electric field inside a capacitor. Recall also that in general, LaTeX: V=-\int{\vec{E}\cdot\vec{dl}}V = − ∫ E → ⋅ d l →. In regions of uniform field, this simplifies to LaTeX: V=|\vec{E}|dV = | E → | d, where d is the distance traveled parallel to the electric field. A capacitor with a gap of 0.5 mm has a potential difference from one plate to another of 22 V. What is the magnitude of the electric field within the capacitor? Give your answer in units of kV/m.

E = 44 kV / m

Explanation:

In a capacitor the potential difference is

         ΔV = -E x

         E = - ΔV / x

Let's reduce to the SI system

         x = 0.5 mm = 0.5 10⁻³ m

Let's calculate

         E = 22 / 0.5 10⁻³

         E = 44 10³ V / m

         E = 44 kV / m

Note that this electric field is uniform

answer: b.

explanation:

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards