Two flat conductors are placed with their inner faces separated by 8 mm. If the surface charge density on one of the inner faces is 33 pC/m2 and the other inner face −33 pC/m2, what is the magnitude of the electric potential difference between the two conductors? Answer in units of V

= 0.0298V

Explanation:

Electric field between two conductor = σ / ε₀

σ = 33pC/m²

= 33 × 10⁻¹²C/m²

ε₀ = 8.85 × 10 ⁻¹²C²/Nm²

E = 33 × 10⁻¹² /  8.85 × 10 ⁻¹²

  = 3.7288N/C

potential difference between two conductors = Ed

where d = 8mm = 8 × 10 ⁻³m

V = 3.7288 ×  8 × 10 ⁻³

   = 29.83 × 10 ⁻³

    = 0.0298V

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

a longitudinal wave moves energy that is to the direction of the wave’s movement. - true