Two point charges (q1 = .45pC and q2 = 8.6 pC) are fixed along the x- axis, sepa

Question

Two point charges (q1 = .45pC and q2 = 8.6 pC) are fixed along the x- axis, separated by a distance d- 8.4 cm. Point P is located at (x,y) (d,d. 1) What is ExP, the value of thex-component of the electric field produced by q1 and q2 at point P? cSubmit You currently have 0 submissions for this question. Only 10 submission are allowed. Hou can make 10 more submissions for this question. 2) What is EPvalue of the y-component of the electric field produced by q1 and 42 at point P? C Submit rou currently have O submissions for this question. Only 10 submission are allowed. ou con make 10 more submissions for this question. 3) A third point charge q3 2.6 uC is now positioned along the y-axis at a distance d 8.4 cm from q1 as shown. What is Ex(P), thex-component of the field produced by all 3 charges at point P? NiC Submrt You currently have O submissions for this question. Only 10 submission are allowed You can make 10 more submissions for this question 4) suppose all charges are now doubled (i.e., q1--9 c.q2-17.2 K, q3-5.2 how will the electric field at p change? Its magnitude will increase by less than a factor of two and its direction will remain the same Its magnitude will increase by less than a factor of two and its direction will change Its magnitude will double and its direction will remain the same its magnitude will double and its direction will change rou currently have 0 submissions for this question. Only 10 submission are allowed. ou can make 10 more submissions for this question. 5) How would you change q1 (keeping q and q3 fixed) in order to make the electric field at point P equal to zero? Increase its magnitude and change its sign Decrease its magnitude and change its sign Increase its magnitude and keep its sign the same Decrease its magnitude and keep its sign the same There is no change you can make to q1 that will result in the electric field at point P being equal to zero. You currently have 0 submissions for this question. Only 10 submission are allowed You can make 10 more submisslons for this question

Explanation / Answer

1 ) Ex = kq1 cos 45 degree /(2 d^2)

= 9e9*-4.5e-6*0.707/(2*0.084^2)

= - 2029018 N/C

2) Ey = kq1 sin 45 degree/(2d^2) + kq2/d^2

= - 2029018 +9e9*8.6e-6/0.084^2

= 8940370 N/C

3) Ex = - 2029018 + 9e9*2.6e-6/0.084^2

= 1287309 N/C

4) third option is correct, magnitude will double, direction same.

5) Fifth option is correct. We can not make electric field zero because whatever changes we do, it effects both x and y component equally.

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