I am having trouble with this problem. The electric field at thepoint x = 5.00 c

Question

I am having trouble with this problem. The electric field at thepoint x = 5.00 cm andy = 0 points in the positive x direction with amagnitude of 15.0 N/C. At the pointx = 15.0 cm and y =0 the electric field points in the positive x directionwith a magnitude of 30.0 N/C. Assumethat this electric field is produced by a single, point charge. (a) Find the location of the point charge.
(x, y) = ( , ) cm

(b) Find the sign and magnitude of its charge.
C (a) Find the location of the point charge.
(x, y) = ( , ) cm

(b) Find the sign and magnitude of its charge.
C

Explanation / Answer

(a) As the electric field is due a single point charge and isalong the positive charge, therefore the charge is placed onx-axis.       So y coordinate iszero       Here the electric filed at15cm is greater than the electric field at 5cm. Therefore thecharge is near to x = 15cm       Let the charge is at adistance of x from the 15cm point. Then it is x + 10m from the 5cmpoint.       then we can haveE15cm = kq/x2                                Then kq/x2 = 30N/C                               orq = (x2/k)(30N/C)       Also we can haveE5cm = kq/(x + 10cm)2                            or15N/C = k[(x2/k)(30N/C)]/(x + 10cm)2                                  [(x+ 10cm)/x]2 = 2                               orx + 10cm = (2)x                               or0.414 x = 10cm                               Thenx = 10cm/0.414                                         = 24.15cm Therefore the charge is at a point (24.15cm, 0) As the cahrge is beyond the points of our observations,therefore the sign of the charge is negative.          Magnitude ofcharge is q =((24.15*10-2m)2/(9*109Nm2/C2))(30N/C)                                               = 1944.075*10-13C                                               = 0.2nC                                Then kq/x2 = 30N/C                               orq = (x2/k)(30N/C)       Also we can haveE5cm = kq/(x + 10cm)2                            or15N/C = k[(x2/k)(30N/C)]/(x + 10cm)2                                  [(x+ 10cm)/x]2 = 2                               orx + 10cm = (2)x                               or0.414 x = 10cm                               Thenx = 10cm/0.414                                         = 24.15cm Therefore the charge is at a point (24.15cm, 0) As the cahrge is beyond the points of our observations,therefore the sign of the charge is negative.          Magnitude ofcharge is q =((24.15*10-2m)2/(9*109Nm2/C2))(30N/C)                                               = 1944.075*10-13C                                               = 0.2nC                            or15N/C = k[(x2/k)(30N/C)]/(x + 10cm)2                                  [(x+ 10cm)/x]2 = 2                               orx + 10cm = (2)x                               or0.414 x = 10cm                               Thenx = 10cm/0.414                                         = 24.15cm Therefore the charge is at a point (24.15cm, 0) As the cahrge is beyond the points of our observations,therefore the sign of the charge is negative.          Magnitude ofcharge is q =((24.15*10-2m)2/(9*109Nm2/C2))(30N/C)                                               = 1944.075*10-13C                                               = 0.2nC                                  [(x+ 10cm)/x]2 = 2                               orx + 10cm = (2)x                               or0.414 x = 10cm                               Thenx = 10cm/0.414                                         = 24.15cm Therefore the charge is at a point (24.15cm, 0) As the cahrge is beyond the points of our observations,therefore the sign of the charge is negative.          Magnitude ofcharge is q =((24.15*10-2m)2/(9*109Nm2/C2))(30N/C)                                               = 1944.075*10-13C                                               = 0.2nC

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