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:
A) pbin = 1.535 Kgm/s (+)
B) pbf = 1.696 Kgm/s (-)
C) Δp = 3.3925 Kgm/s
D) Δvr = 10.249 m/s
Explanation:
Given
Mass of the ball: m = 57.5 g = 0.0575 Kg
Initial speed of the ball: vbi = 26.7 m/s
Mass of the racket: M = 331 g = 0.331 Kg
Final speed of the ball: vbf = 29.5 m/s
A) We use the formula
pbin = m*vbi = 0.0575 Kg*26.7 m/s = 1.535 Kgm/s (+)
B) pbf = m*vbf = 0.0575 Kg*29.5 m/s = 1.696 Kgm/s (-)
C) We use the equation
Δp = pbf - pbin = 1.696 Kgm/s - (-1.535 Kgm/s) = 3.3925 Kgm/s
D) Knowing that
Δp = 3.3925 Kgm/s
we can say that
Δp = M*Δvr
⇒ Δvr = Δp / M
⇒ Δvr = 3.3925 Kgm/s / 0.331 Kg
⇒ Δvr = 10.249 m/s
Answer:
A
Explanation:
hope it helped a lot
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Molecules, The movemnet of molecules
