Numerical Response The electric field strength at a distance of 4.00 cm from a 1.20times 10^-5C point charge expressed in scientific notation, is a.bctimes 10^dN/C . The values of a, b , c, and d are __ __ __ and __ (Record all four digits of your answer in the numerical-response section on the answer sheet.) Delta V=(E)/(9)

Here's how to calculate the electric field strength:<br /><br />1. **Formula:** The electric field strength (E) due to a point charge is given by Coulomb's law: E = kQ/r²<br /><br /> Where:<br /> * k is Coulomb's constant (8.99 x 10⁹ Nm²/C²)<br /> * Q is the charge (1.20 x 10⁻⁵ C)<br /> * r is the distance from the charge (4.00 cm = 0.04 m)<br /><br />2. **Calculation:**<br /> E = (8.99 x 10⁹ Nm²/C²) * (1.20 x 10⁻⁵ C) / (0.04 m)²<br /> E = (10.788 x 10⁴ Nm²/C) / (0.0016 m²)<br /> E = 6.7425 x 10⁷ N/C<br /><br />3. **Scientific Notation and Significant Figures:** The question asks for the answer in the format a.bc x 10ᵈ N/C. We need to round our calculated value to three significant figures:<br /><br /> E = 6.74 x 10⁷ N/C<br /><br />4. **Answer:** Therefore, a = 6, b = 7, c = 4, and d = 7.<br /><br />The values to be entered are $\boxed{6747}$. The provided equation $\Delta V=\frac {E}{9}$ is irrelevant to this problem.<br />