Problem #1
Determine electric at a distance of 2.40 x 10-10 m from the core of a hydrogen atom (the nucleus contains a proton)?Answer;
We have seen that the proton charge q = + e = 1.60 x 10-19 C, the distance of the point to the proton, r = 2.40 x 10-10 m; k = 9 x 100 Nm2/C2, so the electric potential at the point is
V = kq/r = (9 x 109 Nm2/C2)(+1,6 x 10-19C)/(2,40 x 10-10 m)
= 6,00 Volt
Problem #2
The charge q = + 2.5 x 10-6 C is placed at the center of the X-Y field. Determine the potential difference VA - VB if: (a) A (5.0) and B at (0.2) cm, and (b) A at (-4.0) cm and B at (2.0) cm.
Answer
(a) Starting position q = 2.5 x 10-6 C shown in figure (a). rA = 5 cm = 0.05 m; rB = 10 cm = 0.02 m, then
VA – VB = kq/rA – kq/rB = = kq(1/rA – 1/rB)
= (9 x 109 Nm2/C2)(+2,5 x 10-6 C)[1/(0,05 m) – 1/(0,02 m)]
= –675 x 103 V
VA – VB = –675 kV
The starting position q = 2.5 x 10-6 C is shown in figure (b). rA = 5 cm = 0.05 m; rB = 10 cm = 0.02 m, then
VA – VB = kq/rA – kq/rB = = kq(1/rA – 1/rB)
= (9 x 109 Nm2/C2)(+2,5 x 10-6 C)[1/(0,05 m) – 1/(0,02 m)]
= –675 x 103 V
VA – VB = –675 kV
Problem #3
Two electric charges which are +4.2 x 10-5 C and –6.0 x 10-5 C separate at a distance of 34 cm. (a) Determine the electric potential at the point that lies in the two lines of charge and is 14 cm apart from the charge –6.0 x 10-5 C, and (b) Where is the location of the two lines of charge that have zero electric potential?
Answer
(a) For example, the point referred to is point P, as shown in figure (a), then the distance P to q2, which is r2 = 14 cm = 14 x 10-2 m, and the distance m P to q1 is r = (34 - 14) cm = 20 cm = 20 x 10-2 m. Electricity potential at P, according to the requirements,
VP = k(q1/r1 + q2/r2)
= (9 x 109 Nm2/C2){[(+4,2 x 10-5 C)/(20 x 10-2 C)] + [(–6,0 x 10-5 C)/14 x 10-2 C]}
V = –2,0 MV
(b) Suppose, point R which has zero potential is located x cm from q1 as shown in figure (b), then r1 = x cm and r2 = (34 - x) cm.
VR = k(q1/r1 + q2/r2) = 0
q1/r1 = – q2/r2
(4,2 x 10-5 C)/(x) = –(–6,0 x 10-5 C)/(34 – x)
4,2/x = 6/(34 – x)
10x = 7(34 – x)
x = 14 cm
So, the location of the point with zero potential is 14 cm from the q1 charge or 20 cm from the q2 charge.
Problem #4
Small balls loaded with +2.0 nC; –2.0 nC; +3.0 nC; and -6.0 nC is placed at the vertices of a square which has a diagonal length of 0.20 m. Calculate the electric potential in the center of the square!
Answer:
Diagonal length x = 0.2 m then the distance of each charge to the square center point is ½ x 0.2 m = 0.1 m.
VP = k(q1/r1 + q2/r2 + q3/r3 +
= k/r (q1 + q2 + q3 + q4)
VP = [(9 x 109 Nm2/C2)/(0,1 m)][+ 2,0 + (–2,0) + 4,0 + (–6,0)] x 10-9 C
VP = –270 volt
VA – VB = kq/rA – kq/rB = = kq(1/rA – 1/rB)
= (9 x 109 Nm2/C2)(+2,5 x 10-6 C)[1/(0,05 m) – 1/(0,02 m)]
= –675 x 103 V
VA – VB = –675 kV
The starting position q = 2.5 x 10-6 C is shown in figure (b). rA = 5 cm = 0.05 m; rB = 10 cm = 0.02 m, then
VA – VB = kq/rA – kq/rB = = kq(1/rA – 1/rB)
= (9 x 109 Nm2/C2)(+2,5 x 10-6 C)[1/(0,05 m) – 1/(0,02 m)]
= –675 x 103 V
VA – VB = –675 kV
Problem #3
Two electric charges which are +4.2 x 10-5 C and –6.0 x 10-5 C separate at a distance of 34 cm. (a) Determine the electric potential at the point that lies in the two lines of charge and is 14 cm apart from the charge –6.0 x 10-5 C, and (b) Where is the location of the two lines of charge that have zero electric potential?
Answer
(a) For example, the point referred to is point P, as shown in figure (a), then the distance P to q2, which is r2 = 14 cm = 14 x 10-2 m, and the distance m P to q1 is r = (34 - 14) cm = 20 cm = 20 x 10-2 m. Electricity potential at P, according to the requirements,
VP = k(q1/r1 + q2/r2)
= (9 x 109 Nm2/C2){[(+4,2 x 10-5 C)/(20 x 10-2 C)] + [(–6,0 x 10-5 C)/14 x 10-2 C]}
V = –2,0 MV
VR = k(q1/r1 + q2/r2) = 0
q1/r1 = – q2/r2
(4,2 x 10-5 C)/(x) = –(–6,0 x 10-5 C)/(34 – x)
4,2/x = 6/(34 – x)
10x = 7(34 – x)
x = 14 cm
So, the location of the point with zero potential is 14 cm from the q1 charge or 20 cm from the q2 charge.
Problem #4
Small balls loaded with +2.0 nC; –2.0 nC; +3.0 nC; and -6.0 nC is placed at the vertices of a square which has a diagonal length of 0.20 m. Calculate the electric potential in the center of the square!
Answer:
Diagonal length x = 0.2 m then the distance of each charge to the square center point is ½ x 0.2 m = 0.1 m.
VP = k(q1/r1 + q2/r2 + q3/r3 +
= k/r (q1 + q2 + q3 + q4)
VP = [(9 x 109 Nm2/C2)/(0,1 m)][+ 2,0 + (–2,0) + 4,0 + (–6,0)] x 10-9 C
VP = –270 volt
Post a Comment for "Electric Potential Problems And Solutions"