Hi, i need help on the following questions. Thank you.
1) The potential at location A is 441 V. A positively charged particle is released there from rest and arrives at location B with a speed vB. The potential at location C is 805 V, and when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2 vB. Find the potential at B.
2) Four identical charges (+1.6 ìC each) are brought from infinity and fixed to a straight line. Each charge is 0.36 m from the next. Determine the electric potential energy of this group.
3) Identical point charges of +1.7 ìC are fixed to diagonally opposite corners of a square. A third charge is then fixed at the center of the square, such that it causes the potentials at the empty corners to change signs without changing magnitudes. Find the third charge.
Please post questions one at a time.
1). Assume the mass of the particle released from either A or C is the same. Call it M.
(1/2)M vB^2 = 441 - VB
(1/2)M (2vB)^2 = 805 - VB
VB is the potential at point B
805 - VB = 4 (441 - VB)
Solve for VB
To solve question 1, we have two equations:
(1/2)M vB^2 = 441 - VB
(1/2)M (2vB)^2 = 805 - VB
First, let's simplify the second equation:
M vB^2 = 805 - VB
Now we can substitute the first equation into the second equation:
M vB^2 = 805 - VB
M (2vB)^2 = 805 - VB
4M vB^2 = 805 - VB
Now we can expand the equation:
4M vB^2 = 805 - VB
4M vB^2 + VB = 805
Combine the terms:
4M vB^2 + VB = 805
Multiply both sides by 4:
16M vB^2 + 4VB = 3220
Now we have a quadratic equation. Let's rearrange it:
16M vB^2 - VB + 3220 = 0
We can solve this quadratic equation using the quadratic formula:
vB = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 16M, b = -1, and c = 3220. Plugging these values into the equation, we get:
vB = (-(-1) ± √((-1)^2 - 4 * 16M * 3220)) / (2 * 16M)
Simplifying further:
vB = (1 ± √(1 - 20480M)) / (32M)
Now we have the expression for vB. However, in order to find the potential at point B (VB), we need to substitute this expression back into one of the original equations:
(1/2)M vB^2 = 441 - VB
Let's use the second equation:
(1/2)M (2vB)^2 = 805 - VB
Substituting vB = (1 ± √(1 - 20480M)) / (32M), we get:
(1/2)M (2((1 ± √(1 - 20480M)) / (32M)))^2 = 805 - VB
Simplifying further:
(1/2)M ((1 ± √(1 - 20480M)) / (16M))^2 = 805 - VB
(1/2)M (1 ± √(1 - 20480M))^2 / (256M^2) = 805 - VB
Now we can solve for VB. Therefore, VB can be found by evaluating the above equation and solving for it.