A +25 μC test charge experiences force 0.500 N. Find electric field magnitude at its position.

Solution & explanation

1. Given: q = 25 μC; F = 0.500 N.

2. Micro means 10⁻⁶.

   $$q=\frac{25}{1000000}$$

3. Field is force per unit positive charge.

   $$E=\frac{F}{q}$$

4. Substitute SI values.

   $$E=\frac{0.5}{\frac{25}{1000000}}$$

5. First divide the coefficients, then apply the power of ten.

   $$E=0.02\times {10}^6=20000$$

6. State the result with units: E = 2.0×10⁴ N C⁻¹, in the force direction.

E = 2.0×10⁴ N C⁻¹