1. **State the problem:** A charge of 4 microcoulombs (4uC) exerts a force of $3.6 \times 10^{-4}$ newtons to the right on a positive test charge. We need to find the magnitude and direction of the electric field at the location of the test charge. 2. **Formula used:** The electric field $E$ at a point is defined as the force $F$ experienced by a positive test charge $q$ divided by the magnitude of that charge: $$E = \frac{F}{q}$$ 3. **Important rules:** - The direction of the electric field is the direction of the force on a positive test charge. - Units of electric field are newtons per coulomb (N/C). 4. **Given values:** - Force $F = 3.6 \times 10^{-4}$ N (to the right) - Charge $q = 4$ microcoulombs $= 4 \times 10^{-6}$ C 5. **Calculate the magnitude of the electric field:** $$E = \frac{3.6 \times 10^{-4}}{4 \times 10^{-6}} = 90 \text{ N/C}$$ 6. **Direction:** Since the force is to the right on a positive test charge, the electric field direction is also to the right. **Final answer:** The magnitude of the electric field is $90$ N/C, directed to the right.