The total electric potential energy is [tex]\frac{kq_{1} q_{3} }{r_{13} } + \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1} q_{2} }{r_{12} }[/tex].
Electric Potential Energy of a System of Charges :
The system's electric potential energy is equal to the amount of work necessary to create a system of charges by guiding them toward their designated locations from infinity against the electrostatic force without accelerating them. The symbol for it is U.U=W=qV. Electrostatic fields are conservative, therefore the work is independent of the path.
Assume three charges q₁ , q₂ and q₃ bring from infinity to point P.
To bring q₁ no work is done,
[tex]V_{p} = \frac{kq_{1} }{r_{1} }[/tex]
where, V = electric potential energy.
q = point charge.
r = distance between any point around the charge to the point charge.
k = Coulomb constant; k = 9.0 × 109 N.
Now bring q₂,
[tex]V_{2} = \frac{kq_{2} }{r_{2} }[/tex]
Work done by q₁ ;
[tex]W_{1} = V_{p} q_{2} = \frac{kq_{1}q_{2} }{r_{12} }[/tex]
Now bring q₃,
[tex]V_{3} = \frac{kq_{3} }{r_{3} }[/tex]
Work done on q₃ by q₁ and q₂
[tex]W= q_{3} [ V_{1} + V_{2} ][/tex]
[tex]=\frac{kq_{1} q_{3} }{r_{13} }[/tex][tex]+ \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1}q_{3} }{r_{12} }[/tex]
This work done is stored in the form of potential energy.
∴U=W= potential energy of three systems.
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The total electric potential energy is [tex]\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}[/tex]
Electric Potential Energy of a System of Charges
The labor required to establish a system of charges by guiding them toward their intended positions from infinity against the electrostatic force without accelerating them is equal to the electric potential energy of the system. Its identifier is U.U=W=qV. Due to the conservatism of electrostatic fields, the work is independent of the path.
Consider having three charges. Q1, Q2, and Q3 bring point P from infinity.
No work has been done to bring q1,
[tex]V_{1} = \frac{kq1}{r1}[/tex]
where, V = electric potential energy.
q = point charge.
r = distance between any point around the charge to the point charge.
k = Coulomb constant; k = 9.0 × 109 N.
Now bring q₂,
[tex]V_{2} = \frac{kq2}{r2}[/tex]
Work done by q₁ ;
W1 = [tex]V_{p} q2[/tex] = [tex]\frac{kq1q2}{r12}[/tex]
Now bring q₃,
[tex]V_{3} = \frac{kq3}{r3}[/tex]
Work done on q₃ by q₁ and q₂
W= q3{[tex]V_{1} + V_{2}[/tex]}
W = [tex]\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}[/tex]
This work done is stored in the form of potential energy.
∴U=W= potential energy of three systems.
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6. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (3pt) a) How long are her feet in the air? b) What is her highest point above the board? c) What is her velocity when her feet hit the water?
The results of the calculation are;
a) The feet spends 0.41 s in air
b) The highest point above board is 2.62 m
c) The velocity when her feet hit the water is 7.2 m/s
What is the time spent in air?From the data presented;
v = u + at
But v = 0 m/s at the maximum height thus;
0 = 4 - (9.8 * t)
4 = 9.8 * t
t = 4/9.8
t = 0.41 s
b) from;
h = ut - 1/2gt^2
h = (4 * 0.41) - (0.5 * 9.8 * (0.41)^2)
h = 1.64 - 0.82
h = 0.82 m
The total height above board = 0.82 m + 1.8 m = 2.62 m
c) The total time in air is obtained from;
h = ut + 1/2gt^2
u = 0m/s because she dropped off the board
h = 1/2gt^2
2.62 = 0.5 * 9.8 * t^2
t = √2.62/0.5 * 9.8
t = 0.73 seconds
Hence, the velocity when her feet hit the water is obtained from;
v = u + gt
when u = 0 m/s
v = gt
v = 9.8 * 0.73 s
v = 7.2 m/s
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