Question Solved1 Answer Exercise 20.27 6 of 14 > In the figure, point P is midway between the two charges. (Figure 1) Review Constants Find the electric field in the plane of the page 5.0 cm to the left of P. Express your answers in newtons per coulomb. Enter your answers numerically separated by a comma. Assume that the positive r and y axes are directed to the right and upward respectively. ΤΕΙ ΑΣφ ? Ev E = 2.88. 10²,0 N/C Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining Figure < 1 of 1 Part B + 2.0 uC Find the electric field in the plane of the page 5.0 cm directly above P. Express your answers in newtons per coulomb. Enter your answers numerically separated by a comma. Assume that the x and y axes are directed to the right and upward respectively. Р — 2.0 uC VO AE ? -5.0 cm E21,E3y = 2.88 • 10²,0 N/C Submit Previous Answers Request Answer Exercise 20.27 6 of 14 A Review Constants In the figure, point P is midway between the two charges. (Figure 1) ΤΕΙ ΑΣΦ ? E3, E3, = 2.88 • 10?,0 N/C Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Part C Figure < 1 of 1 > Find the electric field in the plane of the page at P. Express your answers in newtons per coulomb. Enter your answers numerically separated by a comma. Assume that the x and y axes are directed to the right and upward respectively. + 2.0 uC P — 2.0 uC ? EVO AED + E3ܨܕE = N/C -5.0 cm Submit Request Answer Problem 20.48 In (Figure 1), take qı = 69 PC, q2 = -38 C, and q3 = 19 uC. Part A Find the components of the electric force on 23- Enter your answers in newtons to two significant figures separated by a comma. IVO AEO ? N Your submission doesn't have the correct number of answers. Answers should be separated with a comma. No credit lost. Try again. Figure < 1 of 1 > Submit Previous Answers Request Answer y (m) Provide Feedback 2- 093 10 91 92 x (m) 2 3

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Transcribed Image Text: Exercise 20.27 6 of 14 > In the figure, point P is midway between the two charges. (Figure 1) Review Constants Find the electric field in the plane of the page 5.0 cm to the left of P. Express your answers in newtons per coulomb. Enter your answers numerically separated by a comma. Assume that the positive r and y axes are directed to the right and upward respectively. ΤΕΙ ΑΣφ ? Ev E = 2.88. 10²,0 N/C Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining Figure < 1 of 1 Part B + 2.0 uC Find the electric field in the plane of the page 5.0 cm directly above P. Express your answers in newtons per coulomb. Enter your answers numerically separated by a comma. Assume that the x and y axes are directed to the right and upward respectively. Р — 2.0 uC VO AE ? -5.0 cm E21,E3y = 2.88 • 10²,0 N/C Submit Previous Answers Request Answer Exercise 20.27 6 of 14 A Review Constants In the figure, point P is midway between the two charges. (Figure 1) ΤΕΙ ΑΣΦ ? E3, E3, = 2.88 • 10?,0 N/C Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Part C Figure < 1 of 1 > Find the electric field in the plane of the page at P. Express your answers in newtons per coulomb. Enter your answers numerically separated by a comma. Assume that the x and y axes are directed to the right and upward respectively. + 2.0 uC P — 2.0 uC ? EVO AED + E3ܨܕE = N/C -5.0 cm Submit Request Answer Problem 20.48 In (Figure 1), take qı = 69 PC, q2 = -38 C, and q3 = 19 uC. Part A Find the components of the electric force on 23- Enter your answers in newtons to two significant figures separated by a comma. IVO AEO ? N Your submission doesn't have the correct number of answers. Answers should be separated with a comma. No credit lost. Try again. Figure < 1 of 1 > Submit Previous Answers Request Answer y (m) Provide Feedback 2- 093 10 91 92 x (m) 2 3

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Answers   E1x, E1y = -2.56 x 107 N/C ,0   E2x , E2y = 1.04 x 107  N/C ,0    E3x, E3y = 5.76 x 107  N/C, 0   F3x , F3y = 2.20 x 105 , 1.34 x 105   N Electric field at point A, 5.0cm Left of P,{:[E=E_(1)-E_(2)],[=(kq_(1))/(r_(1)^(2))-(kq_(2))/(r_(2)^(2))],[q_(1)=q_(2)=q" (say) "],[:.quad E=kq[(1)/(r_(1)^(2))-(1)/(r_(2)^(2))]],[=9xx10^(9)xx2xx10^(-6)[(1)/((2.5 xx10^(-2))^(2))-(1)/((7.5 xx10^(-2))^(2)]):}],[=18 xx10^(3)[(1)/(6.25 xx10^(-4))-(1)/(56.25 xx10^(-4))]H//c],[=(18 xx10^(3))/(10^(-4))[(1)/(6.25)-(1)/(56.25)]N//c],[=18 xx10^(7)[0.16-0.018]N//c],[=18 xx10^(7)xx0.142],[=2.56 xx10^(7)N//c]:}Part BPoint A is 5cm. directly above P. Electric field will be as shown.At point A, Components E_(1)sin theta and E_(2)sin theta are equal and opposite, so they will cancel out. Het electric field at A,{:[E=E_(1)cos theta+E_(2)cos theta],[=((kq_(1))/(r^(2))+(kq_(2))/(r_(2)))cos theta],[" But "|q_(1)|=|q_(2)|" and "],[r_(1)=r_(2)],[:.quadE^(˙)=(2kq)/(r^(2))cos theta]:}{:[:.E=(2kq)/(r^(2))(AP)/(r)=(2kqAP)/(r^(3))],[" How, "AP=5xx10^(-2)cm],[r=sqrt((2.5 xx10^(-2))^(2)+(5xx10^(-2))^(2))],[r=5.6 ... See the full answer

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