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:
Because nothing is blocking the LIGHT!
Explanation:
Light may still be shining on the whole glass but if it wasn't there light would shine brighter than if it was there.
Answer:
a. same
b. less
c. same
d. same?
Explanation:
the mass will always be the same no matter where it is. the weight however depends on the gravity.
Answer:
Explanation:
ASSUMING that block = sled AND that the rope is parallel to the slope.
The force acting parallel due to the weight is
13.6(9.81)sin35.5 = 77.475 N
The maximum friction force is
(0.45)13.6(9.81)cos35.5 = 48.877 N
If rope tension is T
77.475 - 48.877 < T < 77.475 + 48.877
28.6 N < T < 126 N
28.6 N will occur if the block is on the verge of sliding downhill
126 N will occur if the block is on the verge of sliding uphill
Could be any value between them.
Answer:
Work is the energy transferred to or from an object via the application of force along a displacement.