Question

students are investigating the electric fields created by two charged objects (a and b). both objects are the same distance, d from the small positive test charge placed at point x. which way will the test charge be pushed? toward object a because the electric field of object b is stronger toward object b because the electric field of object a is stronger toward object a because the electric field of object b is weaker toward object b because the electric field of object a is weaker

Explanation:

Step1: Recall the formula for electric field

The electric field \(E=\frac{kq}{r^{2}}\), where \(k\) is a constant, \(q\) is the charge, and \(r\) is the distance. Here, \(r = d\) for both charges.

Step2: Compare the electric fields of object A and object B

For object A, \(q_{A}=+ 2\), so \(E_{A}=\frac{k\times2}{d^{2}}\). For object B, \(q_{B}=+1\), so \(E_{B}=\frac{k\times1}{d^{2}}\). Since \(E_{A}>E_{B}\) (because \(\frac{2k}{d^{2}}>\frac{k}{d^{2}}\)), and like charges repel. The test charge (positive) will be repelled more strongly by object A. So it will be pushed toward object B.

Answer:

toward object B because the electric field of object A is stronger