Question

44 A -4nC are 4 nC are located at (i, 0, 3) ancd and ( repectivety Determine the potential at Pi. -2 4% A chage of S aC is placed at each of the four corners of a Sections 4.7 and 4.8-Electric Potential and 135 Tho point charpes -2nC and O sides f s C i placed ia at the point 3 cm above the center of ths Cakulate the electrical potential p 437 a) A total curge opc is split into two equal charges e square around a circular loop of radius * m. Find the potential intervals around the loo b) It O is split into three equal charges spaced at 120 int find the potential at the innc. ois-2nmC, and Q, = 3 mC are, r the potential at the censter ie frin the limt a 87 located at (a 04).1-2.5, 1), and (3, -4, 6). (a) Find the potential Vi at Pl-1, 1.2). (b) Calculate the potential difference Vio if Q is (1,2,3). 4.39 The potential distribution in free space is given by Calculate the electric field strength at (4, π/4,-1). V = nyt: + 3) V. Find (a) E at (3,4.-6) 4.40 (b) the charge within the cube 0 < x <1,0<y<1,0<z<1. 441 The volume charge density inside an atomic nucleus of radius a is p. where p, is a constant (a) Calculate the total charge (b) Determine E and V outside the nucleus. a2 (c) Determine E and V inside the nucleus. (d) Prove that E is maximum at r = 0.745a.

question 41

Answer 1

According to question: volume charge density rho = rho_0 (1 - v^2)/a^2 a) Now let�s assume a small elemental volume at a distance r from the centre the volume is dV = 4 kv^2 dv ... (i) The amount of charge can be obtained as: Q = integral integral integral^2_0 rho dv ... (ii) According to eqn (i), and eqn (ii) Q = integral integral integral^v_0 rho 4kv^2 dv = integral^a_0 4 k (r^2 - vd)/a^2 dv = 4k (a/3 - a^3/5) [answer] b) we need to determine E and V outside nucleus According to Gauss� law, integral integral^s_0 E^rightarrow dA^rightarrow = Q_enc/t_0 Electric field E = E(r) \r\n Therefore, at a distance r: r greaterthanorequalto a from center E (r) 4 pi r^2 = 1/epsilon_0 4 k (a/3 - a^3/5) rightarrow E(r) = 1/r^2 epsilon_0 (a/3 - a^3/5) As we know, E = -nabla V Hence, V = -integral^r_0 E_dr = -(a/3 - a^3/5)/rt_0 [Answer] c) Now, we need to determine E and V inside nucleus