Charge will decreases.
A parallel plate capacitor when it is fully charged to voltage V is given as:
C = Q/V
The capacitance of parallel plate capacitor with two plates of Area A separated by distance d and no dielectric material between plates is
C = ε₀ A /d
since from above equation it shows C is proportional to Q and also C is inversely proportional to distance d.
So, ATQ when d increases C will decrease which in result decreases charge on the capacitor.
Thus, Charge will decrease.
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Answer:
3.25 × 10^7 m/s
Explanation:
Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)
Ek = 1/2 mv2 = qΔV .................. 1
Given that V is the electron speed in m/s
Charge of electron = 1.60217662 × 10-19 coulombs
Mass of electron = 9.109×10−31 kilograms
ΔV = 3.0kV = 3000V
Make V the subject of the formula in eqaution 1
V = sqr root 2qΔV/m
V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31
V = 3.25 × 10^7 m/s
Here Change in Kinetic Energy
= Work Done by Friction
Therefore, substituting the
given values to the equation, we get
0.5 * m * (vFinal^2 -
vInitial^2) = µ m g * d
Therefore
0.5*( 5.90^2 - Vfinal^2 ) =
0.100*9.8*2.10
Therefore
vfinal = 5.54 m/sec
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Answer:
N / NEWTONS
Explanation:
Named after Isaac Newton, the man who discovered gravity
