Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?
Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.
(a) What is the electric potential at point a due to q1 and q2?
(b) What is the electric potential at point b?
(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?
Answer:
a) the potential is zero at the center .
Explanation:
a) since the two equal-magnitude and oppositely charged particles are equidistant
b)(b) Electric potential at point b, v = Σ kQ/r
r = 5cm = 0.05m
k = 8.99*10^9 N·m²/C²
Q = -2 microcoulomb
v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m
v = -105 324 V
c)workdone = charge * potential
work = -6.00µC * -105324V
work = 0.632 J
Answer:
Explanation:
Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.
1 ) Common acceleration a = force / total mass
= 234 / ( 25 +91 )
= 2.017 m s⁻².
2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂
= mass x acceleration
= 25 x 2.017
= 50.425 N
The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.
3 ) Maximum friction force that is possible between m₁ and m₂
= μ_s m₁g
= .79 x 25 x 9.8
= 193.55 N
Acceleration of m₁
= 193 .55 / 25
= 7.742 m s⁻²
This is the common acceleration in case of maximum tension required
So tension in rope
= ( 25 +91 ) x 7.742
= 898 N
4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁
= μ_k m₁g
= .62 x 25 x 9.8
= 151.9 N
Acceleration of m₁
= 151.9 / 25
= 6.076 m s⁻².
When the magnet is placed on the glass, it is attracted to the iron filings. The pattern of the iron filings shows that the lines of force that make up the magnetic field of the magnet. Also, The lines of force of north and south poles attract each other whereas those of two north poles fend off each other.
Answer:
The force is -1620.73 N.
Explanation:
Given that,
Mass of car = 1000 kg
Velocity = 1 m/s
Distance = 2 m
Angle = 30°
We need to calculate the force
Using formula of work done
Put the value into the formula
Hence, The force is -1620.73 N.
Answer:
the maximum angular speed (in radians per second) of a Blu-ray disc as it rotates is 57.6 m/s
Explanation:
Given information:
diameter of the disc, d = 11 cm, r = 5.5 cm = 0.055 m
angular speed ω = 10000 rev/min = (10000 rev/min)(2π rad/rev)(1/60 min/s)
= 1000π/3 rad/s
to calculate the maximum angular speed we can use the following formula
ω = v/r
v = ωr
= (1000π/3)(0.055)
= 57.6 m/s
