A)Vector Eb
The field due to +Q will be in right, that due to -Q will be in left, so the resusltant field would point in between them, due to symmetry and same magnitude of distance and charges, it will point in the direction indicated by vector Eb.
You have two charges, +Q and -Q that are located on the x-axis, at equal distances...
Two charges,+Q and-Q, are located two meters apart and there is a point along the line that is equidistant (midway) from the two charges as indicated. Which vector best represents the direction of the electric field at that point? EB Ec E -Q +Q EA ΕΑ EB Ec ED none
Two equal and opposite point charges, q and -q are located at: (0, 0, d/2) and (0, 0, -d/2), respectively as shown in the figure below. This arrangement is known as an electric dipole. Use first-order Taylor expansions to show that the total electric field due to this electric dipole at very large distances from the origin as compared to the dipole separation, d, is given approximately by: qd t+q d/2 | θ -d/2 -q
Two equal charges are situated on the x axis at 40.0 cm on either side of the origin as shown in the figure below. Location A is on the perpendicular bisector at a distance of 26.0 cm from the origin on the y axis. (Assume that the +x axis is directed to the right and the +y axis is directed up.) (a) If q1 = -2.75 MuC and q2 = -2.75 MuC, what is the net electric field due to...
Two point charges of equal but opposite magnitudes lie on the x and y axis with their coordinates shown below. At point A, what is the direction of the electric field? 1. (o.3) -9 (4,0) Explain your reasoning for your answer. 2. Shown below are two charges along the same line. In which of the labeled positions is most likely where the electric field going to be zero? +2q +q Explain your reasoning for your answer
on 114 points) Two charges, +Q and-Q, are located two meters apart and there is a point along the line that is equidistant (midway) from the two charges as indicated. Which vector best represents the direction of the electric field at that point? EB Ec Ер o +Q EA O EA EB Ес ED none Question 2 (4 points) ✓ Saved MacBook Pro C G Search or type URL % & 4 5 6 7 8 9 0
Two equal charges are situated on the x axis at 40.0 cm on either side of the origin as shown in the figure below. Location A is on the perpendicular bisector at a distance of 31.0 cm from the origin on the y axis. (Assume that the +x axis is directed to the right and the y axis is directed up.) 1 g2 (a) It 41--3.25 C and 42-3.25 uC, what is the net electric field due to both charges...
Two equal charges are situated on the x axis at 40.0 cm on either side of the origin as shown in the figure below, Location A is on the perpendicular bisector at a distance of 22.0 cm from the origin on the y axis. (Assume that the +x axis is directed to the right and the ty axis is directed up.) 91 42 (a) If 91 =ー4.50 uc and q2 =-4.50 μC, what is the net electric field due to...
Two charges are located on the x axis: q1 = +5.5C at x1 = +5.4 cm, and q2 = +5.5C at x2 = -5.4 cm. Two other charges are located on the y axis: q3 = +3.5C at y3 = +4.8 cm, and q4 = -10C at y4 = +6.4 cm. Find (a) the magnitude and (b) the direction of the net electric field at the origin. please help!!! Two charges are located on the x axis: q1 +5.5 1C...
1. Two point charges, q, and q are fixed in position. a is located at (0, d). qg is located at (0,-d). The value of q, is known, and it is positive. The value of q, is unknown. The value of d is known, and it is positive. Also fixed in position is a uniformly charged line segment of length d. This segment is parallel to the x-axis and its left end is located at (d/2.-d). The total electric field...
Two charges (dipole) of +q = +6.00 μC and −q = −6.00 μC along the y-axis, separated by 3.00 m, as shown in the figure below. Point P is located 4.00 m directly to the right of the positive charge, as shown. The origin is located halfway between the charges. (a) At point P (test point), sketch and label the electric field E+ due to the positive charge +q, and the electric field E - due to the negative charge...