Let dx be an infinitesimal length of rod at x . The charge in this segment is dq dx =λ . The charge dq may be considered to be a point charge. The electric field it produces at point P has only an x component and this component is given by dE dx Lax x = +− 1 4 02 π ε λ bg . The total electric field produced at Pby the whole rod is the integral () ()() 2 0 000000111444 , 44 L L x dx E Lax a La Lax Lq aL a aL a εε ε εε λλ λ ⎛⎞ == = − ⎜⎟− π + ⎝⎠ +− λ1 == − π+ π+ ∫ upon substituting qL λ −= . With q = 4.23 × 10 − 15 C, L =0.0815 m and a = 0.120 m, we obtain 3 1.57 10 N/Cx E − =− ×, or 3| | 1.57 10 N/C x E −=× . (c) The negative sign in xEindicates that the field points in the – x direction, or − 180 ° counterclockwise form the + xaxis. (d) If a is much larger than L , the quantity L + ain the denominator can be approximated by a and the expression for the electric field becomes E qa x =− 4 0 2πε . Since 50 m 0.0815 m, aL== ±the above approximation applies and we have 8 1.52 10 N/C
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