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27 Nov 2019

Hi I need help to solve this problems it would really help me tostudy, hopefully someone can help me with as much details aspossible. (formuas, explanations, etc) thank you in advance!

1. A 50.0 kΩ resistor is connected with a 30.0 μF capacitor and a15.0 V battery with negligible internal resistance. Find the timefor the capacitor to be charged from 0 to 75% of its maximumcharge.

2. A point charge q1 =5.00nC is placed at x=4.00cm,and a second point charge q2=-3.00 nC is placed at x= 6.00cm.

a)What is the magnitude and direction of the electric field dueto the charges, at the origin x= 0cm?

b) What is the magnitude of the electrical potential at theorigin?

c) What would be the magnitude and the direction of the electricforce on a Q= 8.0nC charged particle subsequently place at theorigin?

d) How much work would be required to place Q ate the origin?Assume that the electric potential at infinity is 0.

Thank you!

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

Please I need someone to help me solve these problems:
1a) Here, it says that charge q1=10 nC is at theorigin,charge q2 = 16 nC is on the x axis at x= 4m, andchargeq3 = 20 nC is at a point x=4m and y =4m. I wasasked tofind the magnitude and direction of the resultant forceonq2.
b) A charge q is at x= 2m and a second charge -q is at x=-2m, I wasasked to find (a) the electric field on the x axis at anarbitrarypoint where x>2m, and (b) to find the limiting formwherex>>2m.

2)It says a thin rod of length L and charge Q is uniformlychargedwith linear charge density I. I want to find the electricfield atpoint P, where P is an arbitrary point. Please helpme.
Thanks to everyone.

PLEASE HELP ME WITH MY EXTRA CREDIT HOMEWORK PROBLEM, I NEED ALLTHEEXTRA CREDIT POINTS THAT I CAN GET, THANK YOU FOR YOURTIME.

A charge Q1 = -1.6 x10-6C is fixed onthex-axis at +4m, and a charge Q2 = +9 x10-6Cis fixed on the y- axis at +3.0m.

a) Calculate the magnitude ans direction of the electricfield E atthe origin O due to the two charges. Draw andclearly label thisvector on the coordinate axis.

b) Calculate the electric potential V at thee origin.

A charge Q3 = -4 x 10-6C is brought fromavery distant point by an external force and placed attheorigin.
c) On the axis above, indicate and clearly label the directionofthe force on Q3 at the origin.

d) Calculate the work that had to be done by the externalforce tobring Q3 to the origin from the distant point.

PLEASE HELP! I WILL POST EXTRA POINTS FOR YOU AS I AM APLATINUMMEMBER!

Goal Use the superposition principle to calculate the electricfield due to two point charges.

Problem Charge q1 = 7.00 µC is at the origin, and charge q2 = -5.35µC is on the x-axis, 0.300 m from the origin (Fig. 15.12).

(a) Find the magnitude and direction of the electric field at pointP, which has coordinates (0, 0.400) m.

(b) Find the force on a charge of 2.00 10-8 C placed at P.

(a) Calculate the electric field at P.
Find the magnitude of 1 with Equation 15.6.
E1 = 3.93 105 N/C
Vector 1 is vertical, making an angle of 90° with respect to thepositive x-axis. Use this fact to find its components. E1x = E1cos(90°) = 0
E1y = E1 sin(90°) = 3.93 105 N/C
Next, find the magnitude of 2, again with Equation 15.6.
E2 = 1.92 105 N/C
Obtain the x-component of 2, using the triangle in Figure 15.12 tofind cos ?.
E2x = E2 cos ? = (1.92 105 N/C)(0.600)
E2x = 1.15 105 N/C
Obtain the y-component in the same way, but a minus sign has to beprovided for sin ? because this component is directeddownward.
E2y = E2 sin ? = (1.92 105 N/C)(0.800)
E2y = -1.54 105 N/C
Sum the x-components to get the x-component of the resultantvector. Ex = E1x2x = 0 + 1.15 105 N/C
Ex = 1 N/C
+ E
Sum the y-components to get the y-component of the resultantvector. Ey = E1y2y = 3.93 105 N/C - 1.54 105 N/C
Ey = 2 N/C
+ E
Use the Pythagorean theorem to find the magnitude of the resultantvector.
E = 3 N/C

The inverse tangent function yields the direction of the resultantvector.
? = 4°

(b) Find the force on a charge of 2.00 10-8 C placed at P.
Calculate the magnitude of the force (the direction is the same asthat of because the charge is positive). F = Eq = E(2.00 10-8C)
F = 5 N




(a) Place a charge of -3.60 µC at point P and find the magnitudeand direction of the electric field at the location of q2.

E = N/C



? = ° counterclockwise from the positive x-axis



(b) Find the magnitude and direction of the force on q2.

F = N



? = ° counterclockwise from the positive x-axis



I Don't know how to do this Please help Me. Thank you. The top partis how i figured out the rest i think its the missing informationto solve the bottom part.

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