A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Two electricallycharged spheres are suspended from insulated threads a certain distance from each other. There is a certain amount of electrostatic force between them. Describe specifically (not just increase or decrease) what happens to this force in each of the scenarios below
The charge on one sphere is reduced by half
The charge on both spheres is doubled
The distance between the spheres is increased by a factor of three
The distance between the sphere is decreased to onefourth
The charge of each sphere is doubled and the distance between them is doubled
anonymous
 one year ago
Two electricallycharged spheres are suspended from insulated threads a certain distance from each other. There is a certain amount of electrostatic force between them. Describe specifically (not just increase or decrease) what happens to this force in each of the scenarios below The charge on one sphere is reduced by half The charge on both spheres is doubled The distance between the spheres is increased by a factor of three The distance between the sphere is decreased to onefourth The charge of each sphere is doubled and the distance between them is doubled

This Question is Closed

shamim
 one year ago
Best ResponseYou've already chosen the best response.0Force between two electrically charged objects is F=k(q1*q2/r^2)

shamim
 one year ago
Best ResponseYou've already chosen the best response.0Frm my equation If charge of any 1 sphere is reduced to half then force between them will b half of the previous force

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1let's suppose that Q_1 is the charge on sphere #1, and Q_2 is the charge on sphere #2, then the electrostatic force has the subsequent magnitude: \[F = k\frac{{{Q_1}{Q_2}}}{{{d^2}}}\] where d is the distance between the spheres

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1first scenarios: the new charge on sphere #1 is Q_1/2, so we have: \[{F_1} = k\frac{{\left( {{Q_1}/2} \right){Q_2}}}{{{d^2}}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1second scenario: the new charges are: 2Q_1 and 2Q_2 on the spheres #1 and #2 respectively. The new force is: \[{F_2} = k\frac{{{{\left( {2Q} \right)}_1}\left( {2{Q_2}} \right)}}{{{d^2}}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1third scenario: the new distance is 3d, so the new force is: \[{F_3} = k\frac{{{Q_1}{Q_2}}}{{{{\left( {3d} \right)}^2}}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1fourth scenario: the new distance is d/4, so the electrical force is: \[{F_4} = k\frac{{{Q_1}{Q_2}}}{{{{\left( {d/4} \right)}^2}}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1fifth scenario: the new charges are 2Q_1 and 2Q_2 on the spheres #1 and #2 respectively, furthermore the new distance is 2d, so the electrostatic force is: \[{F_5} = k\frac{{{{\left( {2Q} \right)}_1}\left( {2{Q_2}} \right)}}{{{{\left( {2d} \right)}^2}}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: in each of five scenarios, you have to express the new electrostatic force, namely F_1, F_2, F_3, F_4, and F_5 in terms of the force F

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ie. F_1 is onehalf the magnitude initial electrostatic force, F_2 is four times the magnitude initial electrostatic force, F_3 is oneninth the magnitude initial electrostatic force, F_4 is sixteen times the magnitude initial electrostatic force and F_5 is equal in magnitude to the initial electrostatic force.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.