Since the electric field between the plates is constant, If the two plates are brought closer together, the potential difference between the two plates decreases
The relation between potential difference and the electric field is given by ΔV = E.d
Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.
The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.
The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.
Refer to more about the potential difference here
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Answer:
The work done is "2000 J".
Explanation:
The given values are:
Force,
F = 200 N
Mass,
m = 55 kg
Displacement,
d = 10 m
Now,
The work done will be:
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
<em>Given that:</em>
mass of the ball (m) = 0.5 Kg ,
ball strikes the wall (v₁) = 5 m/s ,
rebounds in opposite direction (v₂) = 2 m/s,
time duration (t) = 0.01 s,
<em> Determine the force (F) = ?</em>
We know that from Newton's II law,
<em>F = m. a</em> Newtons
(velocity acting in opposite direction, so <em>a = ( (v₁ + v₂)/t</em>
= m × (v₁ + v₂)/t
= 0.5 × (5 + 2)/0.01
= 350 N
<em>The force acting up on the ball is 350 N</em>
Answer:
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Explanation:
