Parallel plates consider a very large conducting plate at potential v0 suspended a distance d above a very large grounded plane. find the potential between the plates. the plates are large enough so that they may be considered to be infinite. this means that one can neglect fringing fields.

Explanation:

Since the field lines are parallel and the electric field is uniform between two parallel plates, a test charge would experience the same force of attraction or repulsion no matter where it is located in the field,

I attached an image that could help to understand the representation of the field. The formula used to calculate it is given by,

(1)

If we want to consider the change in Voltage with respect to the position then it would be,

According to the information provided, the potential is and there is a distance d, therefore

(2)

Taking equation (1) we can clear V, to what we have,

Integrating,

Substituting (2)

Where x is the height from the grounded plate.

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

the rearrangement of particles in a physical change. hope this