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:
0.026 V
Explanation:
Given that,
Inductance of the coil, L = 6 mH
The current changes from 0.2 A to 1.5 A in a time interval of 0.3 s
We need to find the magnitude of the average induced emf in the coil during this time interval. The formula for the induced emf is given by :
So, the magnitude of induced emf is 0.026 volts.
Answer:
5.3 × 10^(-8) m
Explanation:
We are given;
Shear force; F = 400 N
Length of cube; L_o = 30 cm = 0.3 m
Shear modulus; S = 2.5 × 10^(10) N/m²
Now,the resulting relative displacement can be gotten from the formula;
F = A × S × Δx/L_o
Where Δx is resulting relative displacement
A is area.
Area of cube = (L_o)² = 0.3² = 0.09
Thus, making Δx the subject, we have;
Δx = (F × L_o)/(A × S)
Plugging in the relevant values;
Δx = (400 × 0.3)/(0.09 × 2.5 × 10^(10))
Δx = 5.3 × 10^(-8) m
No I do not agree. It is because work is done when force acting o a body displace or covers certain displacement in the direction of force applied .
