For Chemistry Tutorials
Introduction
 
Vectors
 
Mechanics
 
Energy Work Power
 
Impulse Momentum
 
Rotational Motion
 
Optics
 
Properties Of Matter
 
Heat Temperature And Thermal Expansion
 
Electrostatics
 
Electric Current
 
Magnetism
 
Waves
 
Exams and Problem Solutions
 
--Vectors Exam and Problem Solutions
 
--Kinematics Exams and Problem Solutions
 
--Dynamics Exams and Problem Solutions
 
--Work Power Energy Exams and Problem Solutions
 
--Impulse Momentum Exams and Problem Solutions
 
--Rotational Motion Exams and Problem Solutions
 
--Optics Exams and Problem Solutions
 
--Properties of Matter Exams and Problem Solutions
 
--Heat Temperature and Thermal Expansion Exams and Problem Solutions
 
--Electrostatics Exams and Problem Solutions
 
--Electrostatics Exam1 and Problem Solutions
 
--Electrostatics Exam2 and Problem Solutions
 
--Electrostatics Exam3 and Problem Solutions
 
--Electric Current Exams and Problem Solutions
 
--Magnetism Exams and Problem Solutions
 
--Waves Exams and Problem Solutions
 
Old Version
 


Menu

Electrostatics Exam1 and Problem Solutions


Electrostatics Exam1 and  Problem Solutions

1. If we touch two spheres to each other, find the final charges of the spheres.

Charge per unit radius is found;

qr=(Q1+Q2)/(r1+r2)

qr=(20-5)q/(2r+r)=5q/r

Charge of first sphere becomes;

Q1=qr.r1=5q/r.2r=10q

Charge of second sphere becomes;

Q2=qr.r2=5q/r.r=5q

2. Positively charged sphere B is placed between two neutral spheres A and C. We cut connection of A and C with ground. If we put A closer to the first electroscope and touch C to the sphere of second electroscope, find the type of charge electroscopes have.

A and C are negatively charged by induction. Thus, leaves of both electroscopes are negatively charged.

3. If force applied by charge placed at point B on A is F, find forces applied by charges C and D on A in terms of F.

Free body diagram of forces is given below;

F=k.Q.Q/d2=k.Q2/d2

FC=k.Q.(-4Q)/(√2d)2=-4k.Q2/2d2=-2.k.Q2/d2=-2F

FD=k.Q.2Q/d2=2k.Q2/d2=2F

4. Find the electric field at point A produced by charges q1 and q2 in terms of k, q and d.

We assume that there is a +q charge at point A while finding electric field at point A.

E1=k.(-4q)/d2=-4k.q/d2

E2=k.(16q)/4d2=4k.q/d2

Resultant electric field at point A is;

Eresultant=E1+E2=-4kq/d2+4kq/d2=0

5. Find electric potential energy produced by Q1, Q2 and Q3 in terms of k.q2/r.

Ep=k.Q1.Q2/r

EP1,2=k.10q.(-12q)/3r=-40k.q2/r

EP1,3=k.10q.5q/4r=25k.q2/2r

EP2,3=k.(-12q).5q/5r=-12k.q2/r

EP=EP1,2+EP1,3+EP2,3

EP=-40k.q2/r+25k.q2/2r-12k.q2/r

EP=-39,5k.q2/r


Author:


© Copyright www.PhysicsTutorials.org, Reproduction in electronic and written form is expressly forbidden without written permission of www.PhysicsTutorials.org. Privacy Policy