Question

The electric field 2.45 cm from a small object points toward the object with a strength of 1.79 x 10⁵ n/c. What is the object's charge (in C, include +/- sign)?

Answer 1

The object's charge is approximately
`-9.53 x 10^-10 C.`

Step-by-step explanation:

To find the object's charge, we can use the equation E = kQ/r², where E is the electric field strength, k is Coulomb's constant
`(9 x 10^9 Nm²/C²),` Q is the object's charge, and r is the distance from the object. In this case, we are given the electric field strength
`(1.79 x 10^5 N/C)` and the distance (2.45 cm). Converting the distance to meters (0.0245 m), we can rearrange the equation to solve for Q: Q = Er²/k.

Plugging in the values, we get
`Q = (1.79 x 10^5 N/C) * (0.0245 m)² / (9 x 10^9 Nm²/C²).` Calculating this expression gives us the object's charge, which is approximately
`-9.53 x 10^-10 C.`