Coulomb's Law Problems and Solutions

1 coulomb s Law Problems and force exerted by a point chargeq1on another point chargeq2located at a distancerawayis given by the following formula F=k|q1q2|r2 rwhere ris a unit vector points that coulomb s law gets only the magnitude of the electric force between two point questions are intended for the college level and are difficult. For simple and more relevantpractice Problems on coulomb s law for the high school level, refer to physics Problems for high school and college students only$ coulomb s Law: Problems and the electric force between two charges of5 10 9 Cand 3 10 8 Cwhichare separated byd= 10.

2 The magnitude of the electrostatic force between two point charges is given byCoulomb s law asF=k|q1q2|d2= 9 109 |(5 10 9)( 3 10 8)|( )2= 135 10 6 Nwhere| |denote the magnitude of the that in coulomb s law force formula, the sign of charges is not included only its absolutevalues must be spheres located at distance ofd= 5 cmattract one another with a force ofF= 3 mN. If one of them has three times more charges than the other, find theelectric force between them?Solution: let one of charges beq1=? and the otherq2= 3q1.

3 Then using coulomb s lawformula and solving for the unknown charges, we haveF=k|q1q2|d23 = 9 109 |q1 3q1| q21=(3 10 3)( )(9 109 3)= 10 14 Page 1 coulomb s Law Problems and square root from both sides givesq1= 10 7 CThus, the magnitude of the charges areq1= C andq2= point chargeq1= 2 Clocated at origin and another point chargeq2= 5 Cison the coordinate(x= 3,y= 4) m.(a) Find the electric force on chargeq1.(b) Is the force attractive or repulsive?Solution: the distance between two point charges is found using distance formula (Pythagoreantheorem) as belowd=p(x2 x1)2+ (y2 y1)2which givesd=p32+ 42= 5 m(a) Now, use coulomb s law formula to find the magnitude of the force between two pointcharges as belowF=k|q1q2|d2= 9 109 |(2 10 6)( 5 10 6)|52= 10 3N(b) coulomb s law gives only the magnitude of the electric force.

4 Being repulsive or attractivedepends on the signs of charges attract and unlike charges repel each , the two charges have opposite signs so the electric force between them is point charges are fixed in place in the right triangle shown below, in whichq1= Candq2= C. What is the magnitude and direction of the electricforce on the+ C(let s call thisq3) charge due to the other two charges?Page 2 coulomb s Law Problems and :First, find the electric force due to each charge on theq3, then use the superpositionprinciple to do the vector sum of the figure below, all forces onq3are sketched.

5 Recall that the like charges repel each otherand unlike charges magnitude of coulomb s forces on the chargeq3is obtained as below F13=k|q1||q3|r213 r13= 9 109 10 6 1 10 6 ( )2(cos x+ sin ( y))= r13is the unit vector (a vector whose length is unity) along the line connecting the twocharges and decomposed as shown in the >0 so the electric field lines are along the line betweenq1andq3and directed awayfromq3. From the geometry we see that sin =810and cos = 102 8210=610. Therefore, F13= ( x+ ( y))= ( x y) NPage 3 coulomb s Law Problems and find the electric force due to F23 F23=k|q2||q3|r223 r23= (9 109) 10 6 1 10 6 102 82 10 4m2( y)= ( y)NTherefore, the resultant force on theq3is F3= F13+ F23= ( x y) + ( y)= ( x y) NThe direction of the net force with thexaxis are determined by tan =|Fy|/|Fx|, so = tan 1 = SinceF3x>0 andF3y<0 , the net force lies in the fourth Pythagorean theorem, its magnitude is also found to be F3 =q( )2+ ( )

6 2= small insulating spheres are attached to silk threads and aligned vertically asshown in the figure. These spheres have equal masses of40 g, and carry chargesq1andq2of equal Cbut opposite spheres are brought intothe positions shown in the figure, with a vertical separation of15 cmbetween that you cannot neglect gravity. What is the tension in the lower threads?Page 4 coulomb s Law Problems and :There are three forces acting onq2. The attractive electrostatic forceFedue toq1,tension force in the thread, and gravity.

7 Thus, its free body diagram is as followsThe system is in equilibrium so the net force on theq2is zero ( Fy)2= 0 Fe T mg= 0 T=k|q1||q2|(15)2 mg T= 9 109 2 10 6 2 10 6 ( )2 ( ) = identical particles, each having charge+q, are fixed at the corners of a squareof sideL. A fifth point charge Q(atPpoint) lies a distancezalong the linePage 5 coulomb s Law Problems and to the plane of the square and passing through the center of thesquare. Determine the force exerted by the other four charges on :Because the magnitude and distance of all charges are equal so consider|F1|=|F2|=|F3|=|F4|=k|qQ|r2By symmetry consideration,Fx=Fy= 0.

8 So the direction of one of the forces is: F1z=k|qQ|r2cos k = k|qQ|r3z kWhere we have used from the geometry of the problem cos =z/r. By symmetry F1z= F2z= F3z= F4z= k|qQ|r3z kSo Fz= 4i=1 Fiz= 4kqQr3z kPage 6 coulomb s Law Problems and terms of the parameters of the square and using the Pythagorean theorem, we have:x= 22L , r=vuutz2+ 22L!2 Fz= 4kQq z2+ 22L 2 32z point charges are located at the corners of an equilateral triangle an inthe figure. Find the magnitude and direction of the net electric force on the7 :Same as the previous problem , first we must calculate each of the electric forcesdue to the 2 C, 4 C charges exerted on the third charge then use the superposition principleto determine the net electric force on 7 coulomb s Law Problems and F21=k|q1q2|r212 r21= 9 1092 10 6 7 10 6( )2 r21= r21N r21is the unit vector points fromq2towardq1so if one decomposes it, we get F21= 12 x+ 32 y!

9 N(Notation:F12is the force exerted by point chargeq1on point chargeq2) F31=k|q1q3|r213 r31= 9 109 7 10 6 ( 4) 10 6 ( )2(cos 60 x+ sin 60 ( y)) N= 12 x+ 32( y)!NUsing superposition principle: F1= F31+ F32, we obtain F1= 12 x+ 32 y!+ 12 x+ 32( y)!= x y(N)And its magnitude is F1 =q( )2+ ( )2= NAnd also the direction of the resultant force with the horizontal axis (x) is = tan 1 | || | = SinceF1x>0 andF1y<0 so the net force lies in the fourth point charges are at the corners of a square. The distance from each cornerto the center m.

10 At the center, there is a qpoint charge. What is themagnitude of the net force on this charge?Page 8 coulomb s Law Problems and :Note: the electric force vector between two point charges located at distancerfromeach other is F=k|q1q2|r2 there is a system of point charges and we want to find the net force on one of the charges,we must use the superposition principle the vector sum of the individual electric forces onthe desired charge: Fnet= F1+ F2+ F3+..So we must vector sum the individual forces dueto four point charges on the qin the drawing below shows the direction of the individual forces.