Question

A +9 B Four small spheres (A, B, C, and D), each of mass m=0.02g, are arranged as shown on the figure, all located a distance d=3.00cm from a point P. Spheres A, B, and C each carry a charge of +5.00nC, while sphere D carries a charge of -5.00nC. d +9 d a) When all the spheres are fixed at their positions, -9 what is the electric field E and electric potential V at point P? D b) What is the total electric potential energy of the +9 system (all four spheres) at that time? c) Sphere D is then released from its initial position (while the other spheres remain fixed), and begins to move toward point P. What is its speed at the moment it arrives at point P? #C

Answer 1

Answer ta EA .d. - Ep 1 Fc •cl. * EB 3x10 mi A. -q tq B с qc 5 nc 58 10 c -50c - 5 X 40 0.029 d 3 cm 0.03 m ta Griveni NA -9 z wie ad z -2 2 Ilhen all spheres are fixed at their positions. Electric field due to point charges point charges will be towards the point p. Electric field due to negative charge will be away from P. 16

the figure, the q from magnitude of electric fields will be l&al= (EB) = (Fel='Eol 471&o d² Also from figure; - E. so they cancel out. TI .. Resultant electric field; EB + + E, q 2.X 요 4T Eo d² -9 2 X 9 X 10' x 5 x 10 १ (3x103) ? 1x 105 N/c Electric Р potential at V VA + VB + Vc + Vo Algebraic sum m) q f a 28 4 TICO d: d' d d 9x109 29 ( ( d >9 9 (10' x 2 x 5X10 z 9x109 3x10-2 3 3x 10² v b) From triangle ; APD ADE Ap²+ DP² d² +d² = 12d Similarly AB B C : C D E AD = rad. And; Ac 2d and BD z 2d :

Potal electric potential energy will be O i UAB T UBc + Uco + UAD + UAC + BD - q.a q.q q.(-9) q-(-a) + + + 41 Co AB Be CD AD 9.9 q(-a) + Ас Вр 2 q 1 ji + -1 rad rad - - 41 Eo rad 2d ad ad A oll c) klben D is moved to wards poin't p and reaches p. to qa, qe and qc Electric potential at D due va Vol 4 Vc - 1 A a q a 411E AD BD . СО - + 9x109 x 5x109 8 x 10-2 r2 2 ra 2 15X10 + 2 3 208T X 10° v ENTRU

Electric potential at p. Np Il VA va" Ve+ + ㅗ + + men AP BP CP 9 -9 10' x 5 X10 E 9x409 x 5x 10 -9 x 3 3x10-2 2 45 x 10 4. 5x10² v Change AU in potential energy? 9 Vp - V.) 5*10* ( 4500 - 2871) 8.145 x 10 J This change in potential energy will be potential energy the kinetic energy ka -6 I mrez 8.145 x 10 2 => -6 2x 80145 x 10 0.02 V = 0.03 mis