For the body to be at rest on the incline, all the forces acting on it should balance out each other. Comparing forces along the incline , qE cos 30= \\frac{490}{\\sqrt{3}} \ imes \\frac{\\sqrt{3}}{2} \ imes 0.01 = 2.45 N mg sin 30 = 1 \ imes 9.8 \ imes \\frac{1}{2} = 4.9 N \ herefore qE cos 30 < mg sin 30 Hence , motion is downwards the plane and f is up the inclined plane .Force balance across the vertical direction gives mg = f sin \ heta + N cos \ heta where \ heta = {30}^{o} Similarly, balancing forces along the x direction, N sin \ heta = f cos \ heta + qE Now, if \\mu is the coefficient of friction, then f = \\mu N Substituting the above values in the equations and solving them simultaneously, we get the value of \\mu as \\frac{\\sqrt{3}}{7}