Two charges, q 1 and q 2 , are located atthe origin and at (0.50m, 0) respective

Question

Two charges, q1 and q2, are located atthe origin and at (0.50m, 0) respectively. Where on the x-axis musta third charge, q3, of arbitrary sign be placed to be inelectrostatic equilibrium if a.) q1 and q2 are like charges of equalmagnitude b.) q1 and q2 are unlike charges ofequal magnitude c.) q1 = +3.0C and q2 =-7.0C Two charges, q1 and q2, are located atthe origin and at (0.50m, 0) respectively. Where on the x-axis musta third charge, q3, of arbitrary sign be placed to be inelectrostatic equilibrium if a.) q1 and q2 are like charges of equalmagnitude b.) q1 and q2 are unlike charges ofequal magnitude c.) q1 = +3.0C and q2 =-7.0C

Explanation / Answer

a) Here q1= q2 and are likecharges.     Now unlike charge q3 is placedin betwwen the q1 & q2     so that, distance between q1& q3 = 'x' m and q3 &q2=(0.5-x)m    Force between q1 & q3 ,F1 = q1q3/kx2     .............(1) Force between q3 & q2 ,F2= q3q2/k(0.5 -x)2 .......(2) For electrostatic equlibrium,F1=F2                        Hence,from equations 1 and2 ,we haveq1q3/kx2= q3q2/k(0.5- x)2   
                                                           (0.5-x)2 =x2  (since,q1=q2)                                                                            0.25-x+x2 = x2 or   x=0.25m                Therefore, on x-axis at the point (0.25,0) system is inequilibrium.                                                                        0.25-x+x2 = x2 or   x=0.25m                Therefore, on x-axis at the point (0.25,0) system is inequilibrium. b)Here  q1= q2 and are un-likecharges.(say q1 is positive and q2 isnegative) Now, a positive charge of q3 is placed on thex-axis at a distance "x" m from the charge q2, so thatelectrosattic equilibrium is attained.     so that, distance between q1& q3 = (0.5+x) m and q3&q2= (x)m    Force between q1 & q3 ,F1 = q1q3/k(0.5+x)2     .............(3) Force between q3 & q2 ,F2= q3q2/kx2                    .......(4) For electrostatic equlibrium,F1=F2                        Hence,from equations 3 and4 ,we haveq1q3/k(0.5+x)2= q3q2/kx2   
                                                           (0.5+x)2 =x2  (since,q1=q2)                                                                            0.25+x+x2 = x2 or   x= -0.25m                Therefore, on x-axis at the point (-0.25,0) system is inequilibrium. c)   Now,q1 = +3.0C andq2 = -7.0C Here  q1= q2 and are un-likecharges. Now, a positive charge of q3 is placed on thex-axis at a distance "x" m from the charge q2, so thatelectrosattic equilibrium is attained.     so that, distance between q1& q3 = (0.5+x) m and q3&q2= (x)m    Force between q1 & q3 ,F1 = q1q3/k(0.5+x)2  =3q3 /k(0.5+x)2  .............(5) Force between q3 & q2 ,F2= q3q2/kx2= 7q3 /kx2                        .......(6)    For electrostatic equlibrium,F1=F2                        Hence,from equations 5 and 6 ,we have3q3/k(0.5+x)2= 7q3/kx2   
                                                          3/ (0.5+x)2 =7/x2                                                                              1.25+7x+7x2 = 3x2 or   x=-0.30m or -1.447m                Therefore, on x-axis at the point (-0.30,0) or (-1.447,0)system isin equilibrium.    Force between q1 & q3 ,F1 = q1q3/k(0.5+x)2     .............(3) Force between q3 & q2 ,F2= q3q2/kx2                    .......(4) For electrostatic equlibrium,F1=F2                        Hence,from equations 3 and4 ,we haveq1q3/k(0.5+x)2= q3q2/kx2   
                                                           (0.5+x)2 =x2  (since,q1=q2)                                                                            0.25+x+x2 = x2 or   x= -0.25m                Therefore, on x-axis at the point (-0.25,0) system is inequilibrium. c)   Now,q1 = +3.0C andq2 = -7.0C Here  q1= q2 and are un-likecharges. Now, a positive charge of q3 is placed on thex-axis at a distance "x" m from the charge q2, so thatelectrosattic equilibrium is attained.     so that, distance between q1& q3 = (0.5+x) m and q3&q2= (x)m    Force between q1 & q3 ,F1 = q1q3/k(0.5+x)2  =3q3 /k(0.5+x)2  .............(5) Force between q3 & q2 ,F2= q3q2/kx2= 7q3 /kx2                        .......(6)    For electrostatic equlibrium,F1=F2                        Hence,from equations 5 and 6 ,we have3q3/k(0.5+x)2= 7q3/kx2   
                                                          3/ (0.5+x)2 =7/x2                                                                              1.25+7x+7x2 = 3x2 or   x=-0.30m or -1.447m                Therefore, on x-axis at the point (-0.30,0) or (-1.447,0)system isin equilibrium.                                                                        0.25+x+x2 = x2 or   x= -0.25m                Therefore, on x-axis at the point (-0.25,0) system is inequilibrium. c)   Now,q1 = +3.0C andq2 = -7.0C Here  q1= q2 and are un-likecharges. Now, a positive charge of q3 is placed on thex-axis at a distance "x" m from the charge q2, so thatelectrosattic equilibrium is attained.     so that, distance between q1& q3 = (0.5+x) m and q3&q2= (x)m    Force between q1 & q3 ,F1 = q1q3/k(0.5+x)2  =3q3 /k(0.5+x)2  .............(5) Force between q3 & q2 ,F2= q3q2/kx2= 7q3 /kx2                        .......(6)    For electrostatic equlibrium,F1=F2                        Hence,from equations 5 and 6 ,we have3q3/k(0.5+x)2= 7q3/kx2   
                                                          3/ (0.5+x)2 =7/x2                                                                              1.25+7x+7x2 = 3x2 or   x=-0.30m or -1.447m                Therefore, on x-axis at the point (-0.30,0) or (-1.447,0)system isin equilibrium. Now, a positive charge of q3 is placed on thex-axis at a distance "x" m from the charge q2, so thatelectrosattic equilibrium is attained.     so that, distance between q1& q3 = (0.5+x) m and q3&q2= (x)m    Force between q1 & q3 ,F1 = q1q3/k(0.5+x)2  =3q3 /k(0.5+x)2  .............(5) Force between q3 & q2 ,F2= q3q2/kx2= 7q3 /kx2                        .......(6)    For electrostatic equlibrium,F1=F2                        Hence,from equations 5 and 6 ,we have3q3/k(0.5+x)2= 7q3/kx2   
                                                          3/ (0.5+x)2 =7/x2                                                                              1.25+7x+7x2 = 3x2 or   x=-0.30m or -1.447m                Therefore, on x-axis at the point (-0.30,0) or (-1.447,0)system isin equilibrium.    Force between q1 & q3 ,F1 = q1q3/k(0.5+x)2  =3q3 /k(0.5+x)2  .............(5) Force between q3 & q2 ,F2= q3q2/kx2= 7q3 /kx2                        .......(6)    For electrostatic equlibrium,F1=F2                        Hence,from equations 5 and 6 ,we have3q3/k(0.5+x)2= 7q3/kx2   
                                                          3/ (0.5+x)2 =7/x2                                                                              1.25+7x+7x2 = 3x2 or   x=-0.30m or -1.447m                Therefore, on x-axis at the point (-0.30,0) or (-1.447,0)system isin equilibrium.                                                                        1.25+7x+7x2 = 3x2 or   x=-0.30m or -1.447m                Therefore, on x-axis at the point (-0.30,0) or (-1.447,0)system isin equilibrium.    Hope this may help you

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