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1. Answer the following questions : 1. The distance between two fixed positive charges 4e and e isl. How should athird charge q be arranged for it to be in equilibrium ? Under what conditionswill the equilibrium of the charge q be stabel and when will it be unstable ?2. Two free positive charges 4e and e are a distance a apart. What change isneeded to acheive equilibrium for the entire system and where should it beplaced ?3. A negative point charge 2e and a positive charge e are fixed at a distance l fromeach other. Where should a positive test charge q be placed on the lineconnecting the charges for it to be in equilibrium ? What is the nature ofequilibrium of the test charge with respect to longitudinal motions. Plot thedependence of the force acting on this charge on the distance between it and thecharge te.A charged particle is free to move in an electric field. It will travel5. Two point charges Q and -3Q are placed at some distance apart. If the electricfield at the location of Q is E then at the locality of 3Q, it is |
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Answer» Explanation:3.Let the charge Q be placed at distance x from charge 2q. Now for the charge to be in equilibrium the forces due to both charges must cancel each other or the forces must be equal and opposite. Now the forces due to each charges is given by F 2q =2kQq/x 2 and F q =kQq/(l−x) 2 for equilibrium we must have F 2q =F q or, 2kQq/x 2 =kQq/(l−x) 2 or, (l−x) 2 x 2 =2 or, l−x x = 2 or, x= 1+ 2 2 l l−x x =− 2 or, x= 2 −1 2 l Since second solution of x lies outside the l we will neglect it. Hence the answer is- $x= 2 +1 2 l $ Now if the charge is displaced longitudinally towards charge q then x will increase hence F 2q will decrease and F q will increase. Hence the charge will be forced to move towards charge q. Similarly if it is moved towards charge 2q, F 2q will increase and F q will decrease thus the charge Q will be moved towards charge 2q. Hence the charge Q is in unstable equilibrium with respect to longitudinal motions.2.Let"R" be the distance from the charge q where Q is in equilibrium .Total Force acting on Charge q and 4q:F=kqQ/r² + k4qQ/(l-r)²For Q to be in equilibrium , F should be equated to zero.kqQ/r² + k4qQ/(l-r)²=0(l-r)²=4r²⇒l-r=2r⇒l=3r⇒r=l/3Taking the third charge to be -Q (say) and then on applying the condition of equilibrium on + q chargekQ/(L/3)² =k(4q)/L²9kQ/L²=4kq/L²9Q=4qQ=4q/9Therefore a point charge -4q/9 should be placed at a distance of L/3 RIGHTWARDS from the point charge +q on the line joining the 2 charges.1.Given,qA = +4eqB = +eAB = xLet, Q be the third point charge located at some point P.AP = xtherefore, PB = (a-x)Force on Q due to qA,F1 = kqAQ/x2Force on Q due to qBF2= kQqB/(a-x)2Since the charge is in equilibrium,F1 = F2kqAQ/x2 = k qBQ/(a-x)24e/x2 = e/(a-x)24/x2 = 1/(a-x)2 (Taking square ROOT on both sides)2/x = 1/(a-x)2(a-x) = x(1)2a - 2x = x2a = 3xx = 2/3aTherefore, the third point charge Q should be placed 2/3a from A.if helps mark as brainlist answer |
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