Question

What is the electric force from the rod on the point charge Q? Suppose that Q-13.1 nC and L = 4.07 cm. Break the rod up into four equal pieces and use superposition to find the force 2 mC 4L

Answer 1

to the right end of the rodis, The distance from the charge di = 14 +(32)* = VIOL The distance from the charge dx = +(L)* = V2L to the left end of the rodis, to the rodis, The distance from the point charge r=L From the figure, sin 2 = 20 = sin & = sine - Fac. = sin 8= The Inear charge density is, 2=9 42 The electric field due to charged rod along the y-axis is, E, - (sin &+sin @) (4) / 3.1 E, - deydi 1,=() Case E, = 4.498x10' NIC

From the figure, 2018 – 2008 - cose, - ਝ = cose, -ਤੇੜ The Inear charge density is, The electric field due to charged rod along the x-axis is, E, = 4 (cosਥ -s cost) 47ਣpr E,=ਣ, ਉo ) 4. = 9 10) ਉਰ ਤੇ) ੬, =-1.0618x10 {C

The electric force along the x-axis is, F, = QE, = (13.1x10°)(-1.0618x10°)=-13.91 N The electric force along the y-axis is, F, = QE, = (13.1x10")(4.498x10°) = 58.9238 N The required electric force is, F=(-13.91 N),(58.9238 N)), (ON) F=(-13.91 N,58.9238 NON) F(-13.9 N,58.9 NON) FX(-14 N,59 N,ON)