PART A)
Electrostatic potential at the position of origin is given by

here we have



now we have


Now work done to move another charge from infinite to origin is given by

here we will have

so there is no work required to move an electron from infinite to origin
PART B)
Initial potential energy of electron




Now we know



now by energy conservation we will have
So here initial total energy is sufficient high to reach the origin
PART C)
It will reach the origin
Answer:
<em>1.228 x </em>
<em> mm </em>
<em></em>
Explanation:
diameter of aluminium bar D = 40 mm
diameter of hole d = 30 mm
compressive Load F = 180 kN = 180 x
N
modulus of elasticity E = 85 GN/m^2 = 85 x
Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A =
=
= 1256.8 mm^2
area of hole a =
=
= 549.85 mm^2
Total contraction of the bar =
total contraction =
==>
= <em>1.228 x </em>
<em> mm </em>
Explanation:
Initial energy = final energy + work done by friction
PE = PE + KE + W
mgH = mgh + 1/2 mv² + W
(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000
v = 22.1 m/s
Without friction:
PE = PE + KE
mgH = mgh + 1/2 mv²
(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²
v = 23.4 m/s