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Figure 22-8c is identical to Fig. 22-8a except that particle 3 now lies on the x axis between particles 1 and 2. Particle3 has charge q_(2)= -3.20 xx 10^(-19)C and is at a distance (3//4)R from particle 1. What is the net electrostatic force vecF_("1.net") on particle 1 due to particles 2 and 3? |
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Answer» Solution :KEY IDEA The presence of particle 3 does not alter the electrostatic force on particle 1 from particle 2. THUS, force `vecF_(12)` still acts on particle 1. Similarly, the force `vecF_(13)` that acts on particle 1 DUE to particle 3 is not affected by the presence of particle 2. Because particles 1 and 3 have charge of opposite signs, particle 1 is attracted to particle 3. Thus, force `vecF_(13)` is directed toward particle 3, as indicated in the free-body diagram of Fig. 22-8d. Three particles : To find the magnitude of `vecF_(13)`, we can rewrite Eq. 22-4 as `F_(13)=(1)/(4pi epsilon_(0))(|q_(1)||q_(3)|)/(((3)/(4)R)^(2))` `=(8.99xx10^(9)N*m^(2)//C^(2))xx ((1.60xx10^(-19)C)(3.20xx10^(-19)C))/(((3)/(4))^(2)(0.0200 m)^(2))` `=2.05xx10^(-24)N`. We can also write `vecF_(13)` in unit-VECTOR notation: `vecF_(13)=(2.05xx10^(-24)N)hati`. The net force `vecF_("1,net")` on particle 1 is the vector sum of `vecF_(12)`, and `vecF_(13)`, that is, from Eq. 22-7, we can write the net force `vecF_("1,net")` on particle 1 in unit-vector notation as `vecF_("1,net")=vecF_(12)+vecF_(13)` `=-(1.15xx10^(-24)N)hati+(2.05xx10^(-24)N)hati` `=(9.00xx10^(-25)N)hati`.(ANSWER) Thus, `vecF_("1, net")` has the following magnitude and direction (relative to the positive direction of the x axis): `9.00xx10^(-25)N and 0^(@)`. (Answer) |
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