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
The equator would be warmer than the poles. :)
Lithium-Dull
Sodium-Dull
Potassium- Dull
Rubidium- Dull
Cesium- Both
Francium- Shiny
Answer:
37.125 m
Explanation:
Using the equation of motion
s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration
<u>Distance during acceleration</u>
Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.
Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be
a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}
Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion
s=0*4.5+ 0.5*1*4.5^{2}=0+10.125
=10.125 m
<u>Distance at a constant speed</u>
At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time
Distance=4.5 m/s*6 s=27 m
<u>Total distance</u>
Total=27+10.125=37.125 m
