Problem #1
The 5 μC point load is shifted from 4 m to 2 m which is measured against other point loads –25 mC. Determine the effort needed to move this position.Answer;
It is known that q1 = 5 μC = 5 x 10-6 C, q2 = 25 mC = 25 x 10-3 C, r1 = 4 m and r2 = 2 m, then using the formula
W12 = –kq1q2 (1/r2 - 1/r1)
= –9 x 109 x 5 x 10-6 x 25 x 10-3 1/2 - 1/4)
W12 = 281.25 J
Problem #2
The system given below consists of q, -2q and 4q charges. Determine the total electrical potential energy of the system!
Answer;
From the system above we get the potential energy total
EP = EP12 + EP13 + EP23 with
EP12 = kq1q2/r12 = k(-2q)(q)/3
EP13 = kq1q3/r13 = k(-2q)(4q)/
EP23 = kq2q3/r23 = k(q)(4q)/4x
then,
EPtotal = = –2kq2/3x + (–6kq2/5x) + kq2/x = –13kq2/15x
Problem #3
Determine the amount of effort and electric potential energy to move a charge of 3.0 μC from a point that has a potential of +12 V to a point that has a potential of –20 V?
Answer;
Electric potential energy is given by
ΔEP = q(V2 - V1) = 3.0 x 10-6(–20 - 12) = -96 J
The amount of effort required is equal to
W = –ΔEP = 96 V
Problem #4
In a homogeneous electric field E = 270 N/m, the charge q = 5 μC moves from position A to position B as far as 20 cm parallel to the electric field and then points to point C 300 towards the direction of the electric field as far as 10 cm (Note the image below this). Determine the change in electric potential energy from point (a) A to point B, (b) B to point C and (c) A to point C!
Answer;
Electric field size E = 270 N / m, q = 5 x 10-6 C, charge q = 5 μC = 5 x 10-6 C.
To look for changes in electrical potential energy, we use the concept
ΔEP = –W = –Fd cos θ = –qEd cos θ, then
(a) from A to B
ABEPAB = -qEd cos θ = –5 x 10-6 x 270 x 0.2 m cos 00
ABEPAB = –2.7 x 10-4 J
(b) from B to C
ΔEPBC = -qEd cos θ = –5 x 10-6 x 270 x 0.1 m cos 300
ΔEPBC = -6.75 x 10-5 J
(c) from A to C
ΔEPAC = ΔEPAB + ΔEPBC = –3,375 x 10-5 J
Problem #5
The electric field vector is given by E = (0.5x + 2y) i + 2xj (N / C). Determine the effort to move the charge q = –2μC in this electric field from (a) (0,0,0) to (4,0,0) m, (b) (4,0,0) to (4,2, 0) m and (c) (4,2,0) to (0,0,0) along a straight line.
Answer;
If the charge moves from point A to point B, the effort carried out by the electric field is given by
In this case the integral is calculated along the path connecting points A and B. Then,
(a) We calculate integrals along a straight line that connects the starting point and end point:
Wa = ∫04 (y = 0) qExdx = ∫04 (y = 0) q(0.5x) dx = q (0.5)(0.5)(4)2 = 4q = ─8 x 10-6 J
(b) We calculate integrals along a straight line that connects the starting point and end point:
Wb = ∫02 (x = 4) qEydy = ∫02 (x = 4)q(2)(4)dy = q(8)(2) = 16q = ─3.2 x 10-5 J
(c) We calculate integrals along a straight line that connects the starting point and end point:
Wc = ∫AB (y = x/2)q(Exdx + Eydy) = ∫AB (y = x/2)q[(0.5x + 2y)dx + (2x)dy]
= ∫X0 = 4q[(0.5x + x)dx + (2x)dx/2]
Wc = ∫X0 = 4q(2.5x)dx = ─20q = 4.0 x 10-5 J
Post a Comment for "Electric Potential Energy Problems And Solutions"