a charge q1= 5C is at the origin and a second charge q2= -3C is on the x-axis 80 cm away from the origin find the electric field at a point on the y axis 50 cm away from origin when angle 0=60

Answer 1

An ELECTRIC FIELD is said to exist at any point in space where a charge (q) experiences a force (Fq). The direction of the field at any point is in the same direction that a POSITIVE TEST CHARGE would experience a force. Recall like sign charges repel (+,+ or -,-), and that opposite charges attract (+,-).

The STRENGTH of the electric field at any point is given by the equation E = F/q, where F is given by Coulomb's Law (F = Fq = kq1q2/r^2). Substituting this into the numerator for F, gives E = kq/r^2. Where "k" is Coulomb's constant with a value of 9 x 10^9 with units (Newton)x(meter squared) / (Coulomb's squared). The electric field is a vector so you find the strength of the field from each charge at the "y = 50 cm" location due to each charge separately, then add the vectors to find the resultant E-field.

So a POSITIVE TEST CHARGE at y = 50cm (convert cm to m) will experience a REPULSIVE force due to the same + 5C charge at the origin; F = (9x10^9Nm^2/C^2)(5C)/(.5m)^2 = 1.8 x 10^11N , and an attractive force due to the -3C charge at x = 80cm,and using Pythagorean's theorem to find the distance to the y = 50cm point... F = (9x10^9Nm^2/C^2)(-3C)/(,94m)^2 = -3.1 x 10^10N. Ignore the negative, just know the direction of this force is TOWARD the -3C point charge. The total E-fied is the sum of each; Et = E1 + E2..