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Figure 22-8e is identical to Fig 22-8a except that particle 4 is now included. It has charge q_(4)= -3.20 xx 10^(-19)C, is at a distance (3)/(4) R from particle 1, and lies on a line that makes an angle theta=60^(@) with the x axis. What is the net electrostatic electrostatic electrostatic force vecF_("1, net") on particle 1 due to particle 2 and 4? |
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Answer» Solution :KEY IDEAS The net force `vecF_("1, net")` is the vector sum of `vecF_("1, net")` and a new force `vecF_(14)` acting on PARTICLE 1 due to particle 4. Because particles 1 and 4 have charge of opposite signs, particle 1 is attracted to particle 4. Thus, force `vecF_(14)` on particle 1 is directed toward particle 4, at angle `theta=60^(@)`, as indicated in the free-body diagram of Fig. 22-8f. Four particles: We can rewrite `Eq. 22-4` as `F_(14)=(1)/(4PI epsilon_(0)) (|q_(1)||q_(4)|)/(((3)/(4)R)^(2))` `=(8.99xx10^(9)N*m^(2)//C^(2))xx ((1.60xx10^(-19)C)(3.20xx10^(-19)C))/(((3)/(4))^(2)(0.0200m)^(2))` `=2.05xx10^(-24)N`. Then from Eq. 22-7, we can write the net force on `vecF_("1.net")` particle 1 as `vecF_("1, net")=vecF_(12)+vecF_(14)`. Because the forces `vecF_(12) and vecF_(14)` are not directed ALONG the same axis, we cannot sum simply by combining their magnitudes. Instead, we must add them as vectors, using ONE of the following methods. Method 1.summing directly on a vector-capable calculator. For `vecF_(12)`, we enter the MAGNITUDE `1.15xx10^(-24)` and the angle `180^(@)` For `vecF_(14)`, we enter the magnitude `2.05xx10^(-24)` and the angle `60^(@)`. Then we add the vectors. Method 2. Summing in unit-vector notation. First we rewrite `vecF_(14)` as `vecF_(14)=(F_(14) cos theta) hati+(F_(14) sin theta) hatj`. Substituting `2.05xx10^(-24)N` for `F_(14) and 60^(@)` for `theta`, this becomes `vecF_(14)=(1.025xx10^(-24)N)hati+(1.775xx10^(-24)N)hatj`. Then we sum: `vecF_("1, net")=vecF_(12)+vecF_(14)` `= -(1.15xx10^(-24)N)hati+(1.025xx10^(-24)N)hati+(1.775xx10^(-24)N)hatj` `=( -1.25xx10^(-25)N)hati+(1.78xx10^(-24)N)hatj`. (Answer) Method 3. Summing components axis by axis. The sum of the x components gives us `F_("1, net, x")=F_(12,x)+F_(14,x)=F_(12)+F_(14)cos 60^(@)` `= -1.15xx10^(-24)N+(2.05xx10^(-24)N)(cos60^(@))` `= -1.25xx10^(-25)N`. The sum of the y components gives us `F_("1, net, y")=F_(12,y) +F_(14,y)=0+F_(14)sin60^(@)` `=(2.05xx10^(-24)N) (sin60^(@))` `=1.78xx10^(-24)N`. The net force `vecF_("1, net") ` has the magnitude `F_("1, net")= sqrt(F_("1, net, x")^(2)+F_("1, net, y")^(2))=1.78xx10^(-24)N`. (Answer) To find the direction of `vecF_("1, net")`, we take `theta ="tan"^(-1)(F_("1, net,y"))/(F_("1,net,x"))=-86.0^(@)`. However, this is an unreasonable result because `vecF_("1,net") ` must have a direction between the directions of `vecF_(12)` and `vecF_(14)`. To correct `theta`, we add `180^(@)`, obtaining `=86.0^(@)+180^(@)=94.0^(@)` (Answer) |
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